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NOI
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物理
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<div class="op">
<h1>下拉查看OI错题<br/>NOIP2024 AFO了</h1>
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<br/>
<h1>错题集</h1>
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<div class="t1"><img alt="johnson" onclick="window.open('https:\/\/upload-bbs.miyoushe.com\/upload\/2024\/10\/22\/286158760\/60fcdfbd4e12245f4d69315c48a80154_7170574471892578999.png')" src="https://upload-bbs.miyoushe.com/upload/2024/10/22/286158760/60fcdfbd4e12245f4d69315c48a80154_7170574471892578999.png"/></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P4779">P4779 【模板】单源最短路径（标准版）</a><a href="https://www.luogu.com.cn/problem/list?difficulty=3">普及/提高-</a>不愧是模板题，我也是直接秒了！看好了，板子是这么背的！<a href="https://upload-bbs.miyoushe.com/upload/2024/10/22/286158760/bc91f2048feafc821d94d2184a1939fc_3137352780557072498.png">单源最短路模板</a></div>
<div class="t1"><img alt="" src="https://upload-bbs.miyoushe.com/upload/2024/10/21/286158760/4be9add32577548a072645ce77498592_4896357184329688696.jpg"/>我要省一，呜呜呜！我也是赌上去了！</div>
<div class="t1">AC(Accepted)保佑我NOIP2024 RP++直接拿下400pts！！！<img alt="AC" src="https://upload-bbs.miyoushe.com/upload/2024/10/22/286158760/d4264f3bf787c0d054258a9bddd5c129_6098923047941144383.png"/></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P1038">P1038 [NOIP2003 提高组] 神经网络</a><a href="https://www.luogu.com.cn/problem/list?difficulty=4">普及+/提高</a>这道题真是把我坑惨了，预处理的顺序不一样还会出错。而且输出的是输出层，题目中说了，我没看见。<pre>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
using namespace std;
#define int long long
#define pii pair<int,int>
const int N = 200;
struct Node{
int C,U;
}node[N];
vector<pii> e[N];
int n,p;
int ind[N],outd[N];
signed main(){
cin>>n>>p;
for(int i=1;i<=n;i++){
cin>>node[i].C>>node[i].U;
node[i].C-=node[i].U;
}
for(int i=1,u,v,w;i<=p;i++){
cin>>u>>v>>w;
e[u].push_back({v,w});
ind[v]++;outd[u]++;
}
queue<int> q;
for(int i=1;i<=n;i++){
if(ind[i]==0){
q.push(i);
node[i].C+=node[i].U;
}
}
while(!q.empty()){
int u=q.front();
q.pop();
for(auto ed:e[u]){
int v=ed.first,w=ed.second;
if(node[u].C>0)node[v].C+=node[u].C*w;
ind[v]--;
if(ind[v]==0){
q.push(v);
}
}
}
bool flag=1;
for(int i=1;i<=n;i++){
if((!outd[i])&&node[i].C>0){
cout<<i<<" "<<node[i].C<<'\n';
flag=0;
}
}
if(flag){
cout<<"NULL";
}
return 0;
}
</pre><i>参考资料：<a href="https://blog.csdn.net/Salvator_/article/details/123056485">CSDN题解</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P9228">P9228 原神</a><a href="https://www.luogu.com.cn/problem/list?difficulty=3">普及/提高-</a>大佬的压行真牛逼，这道题是贪心，就是2种情况，详细见题解。<pre>
#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define int long long
const int N = 2e6;
bool cmp(int a,int b){
return a>b;
}
int n,m,k,kk;
int a[N],b[N],ans;
signed main(){
cin>>n>>m>>k;
for(int i=1;i<=n;i++)cin>>a[i],ans+=a[i];
for(int i=1;i<=m;i++)cin>>b[i],ans+=b[i];
sort(a+1,a+n+1,cmp);
int minn=min(n,m);
for(int i=1;i<=minn;i++)if(a[i]>k)ans+=a[i],kk++;
ans+=k*(minn-kk);
return cout<ans,0;
}
</pre><i>参考资料：<a href="https://www.luogu.com.cn/problem/solution/P9228">题解区@封禁用户</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P8881">P8881 懂事时理解原神</a><a href="https://www.luogu.com.cn/problem/list?difficulty=3">普及/提高</a>这道题是判断环的题目，如果有环，那么一定输出0，无环就是1。<pre>
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
const int N = 2e5;
vector<int> e[N];
int t,n,m,ans;
bool vis[N];
void dfs(int u,int fa){
vis[u]=1;
for(auto v:e[u]){
if(v==fa)continue;
if(!vis[v])dfs(v,u);
else ans=0;
if(!ans)return;
}
}
int main(){
cin>>t;
while(t--){
memset(vis,0,sizeof(vis));
ans=1;
cin>>n>>m;
for(int i=1;i<=n;i++){
e[i].clear();
}
for(int i=1,u,v;i<=m;i++){
cin>>u>>v;
e[u].push_back(v);
e[v].push_back(u);
}
dfs(1,0);
cout<<ans<<".000\n";
}
return 0;
}
</pre><i>参考资料：<a href="https://www.luogu.com.cn/problem/solution/P8881">题解区@Manchester_Is_Blue</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P3197">P3197 [HNOI2008] 越狱</a><a href="https://www.luogu.com.cn/problem/list?difficulty=3">普及/提高-</a><p>这道题是组合数学的题目。我们想简单一点，第一个人有m种宗教可以选，第二个人为了和第一个人不同，那么她就只有m-1种宗教可以选，第三种为了和第二种不同，也只有m-1种宗教可以选，以此类推，那么不会冲突的情况就有m*(m-1)<sup>n-1</sup>种。那么总的情况数减去她就是不合法的情况数了，总的情况就是每个人都可以选择m种宗教，最后就是m<sup>n</sup>种。<br/>综上，答案就是m<sup>n</sup>-m*(m-1)<sup>n-1</sup>。</p><pre>
#include <iostream>
#include <algorithm>
using namespace std;
#define int long long
const int mod = 100003;
int n,m;
int pow(int x,int y){
int res=1;
while(y){
if(y&1){
res*=x;
res%=mod;
}
x*=x;
x%=mod;
y>>=1;
}
return res%mod;
}
signed main(){
cin>>m>>n;
cout<<((pow(m,n))-m%mod*(pow((m-1),(n-1)))%mod+mod)%mod;
return 0;
}
</pre>注意加法取模的顺序不同会导致结果不同！只有乘法取模顺序不同才是等价的，加法不等价！<i>参考资料：<a href="https://www.luogu.com.cn/problem/solution/P3197">题解区@蒟蒻初音ミク</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P2024">P2024 [NOI2001] 食物链</a><a href="https://www.luogu.com.cn/problem/list?difficulty=4">普及+/提高</a>这道题很有意思，就是将并查集放大3倍，分为3类：fa[x]表示x本身，fa[x+n]表示x的猎物，fa[x+2*n]表示x的天敌。其中x的猎物的天敌是猎物。总之这个关系有点蒙，画个有向图就好了。总之这个关系怎么表示都可以，最后只要输出正确就行了。<pre>
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 5e5+10;
int n,k;
int fa[N],ans;
int find(int x){
if(x==fa[x])return x;
return fa[x]=find(fa[x]);
}
void merge(int x,int y){
x=find(fa[x]),y=find(fa[y]);
fa[x]=y;
}
int main(){
cin>>n>>k;
for(int i=1;i<=n*3;i++)fa[i]=i;
int opt,x,y;
while(k--){
cin>>opt>>x>>y;
if(x>n||y>n){
ans++;
}
else if(opt==1){
if(find(x+n)==find(y)||find(x+2*n)==find(y)){
ans++;
}
else{
merge(x,y);
merge(x+n,y+n);
merge(x+2*n,y+2*n);
}
}
else{
if(x==y){
ans++;
}
else{
if(find(x+2*n)==find(y)||find(x)==find(y)){
ans++;
}
else{
merge(x+n,y);
merge(y+2*n,x);
merge(x+2*n,y+n);
}
}
}
}
cout<<ans;
return 0;
}
</pre><i>参考资料：<a href="https://www.luogu.com.cn/problem/solution/P2024">题解区@檀黎斗·神</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P3387">P3387 【模板】缩点</a><a href="https://www.luogu.com.cn/problem/list?difficulty=4">普及+/提高</a>隔了一段时间再来写这道题真的受罪匪浅。<pre>
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
#define int long long
const int maxn = 2e6;
vector<int> e[maxn],e2[maxn];
int dfn[maxn],low[maxn],tot;
int stk[maxn],instk[maxn],top;
int scc[maxn],cnt;
int w[maxn],w2[maxn];
int n,m;
void tarjan(int u){
dfn[u]=low[u]=++tot;
stk[++top]=u;instk[u]=1;
for(auto v:e[u]){
if(!dfn[v]){
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(instk[v]){
low[u]=min(low[u],dfn[v]);
}
}
if(low[u]==dfn[u]){
cnt++;
int v;
do{
v=stk[top--];
instk[v]=0;
scc[v]=cnt;
w2[cnt]+=w[v];
}while(v!=u);
}
}
int f[maxn];
signed main(){
cin>>n>>m;
for(int i=1;i<=n;i++){
cin>>w[i];
}
for(int i=1,u,v;i<=m;i++){
cin>>u>>v;
e[u].push_back(v);
}
for(int i=1;i<=n;i++){
if(!dfn[i]){
tarjan(i);
}
}
for(int i=1;i<=n;i++){
for(auto v:e[i]){
if(scc[v]!=scc[i]){
e2[scc[i]].push_back(scc[v]);
}
}
}
for(int i=cnt;i>=1;i--){
if(!f[i]){
f[i]=w2[i];
}
for(auto v:e2[i]){
if(f[v]<f[i]+w2[v]){
f[v]=f[i]+w2[v];
}
}
}
int ans=0;
for(int i=1;i<=cnt;i++){
ans=max(ans,f[i]);
}
cout<<ans;
return 0;
}
</pre><i>参考资料；<a href="https://www.cnblogs.com/avarice/p/16500338.html">博客园题解</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/record/191541429">P11324 【MX-S7-T2】「SMOI-R2」Speaker</a><a href="https://www.luogu.com.cn/problem/list?difficulty=4">普及+/提高</a>这道题是熟练剖分+分类讨论，其实看了题解之后也能看出来了，确实是分类讨论。<pre>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
const int N = 200010, M = N * 2, K = log2(N) + 2;
int n, m;
int h[N], e[M], ne[M], w[M], idx;
LL v[N];
LL depth[N], fa[N][K], maxv[N][K];
LL dist[N], max1[N], max2[N], max3[N];
LL mak1[N], mak2[N], mak3[N];
LL sum[N], res, kk;
void add(int a, int b, int c)
{
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++ ;
}
void dfs1(int u, int fas)
{
max1[u] = v[u];
for (int i = h[u]; i != -1; i = ne[i])
{
int j = e[i];
if (fas == j) continue;
dist[j] = dist[u] + w[i];
dfs1(j, u);
// cout << u << ' ' << max1[u] << endl;
// cout << j << ' ' << max1[j] << endl;
if (max1[j] - 2 * w[i] >= max1[u])
{
max3[u] = max2[u], mak3[u] = mak2[u];
max2[u] = max1[u]; mak2[u] = mak1[u];
max1[u] = max1[j] - 2 * w[i], mak1[u] = j;
// cout << max1[u] << endl;
}
else if (max1[j] - 2 * w[i] >= max2[u])
{
max3[u] = max2[u], mak3[u] = mak2[u];
max2[u] = max1[j] - 2 * w[i], mak2[u] = j;
}
else if (max1[j] - 2 * w[i] >= max3[u])
{
max3[u] = max1[j] - 2 * w[i];
mak3[u] = j;
}
// cout << 'a';
}
}
void dfs2(int u, int fas)
{
depth[u] = depth[fas] + 1;
fa[u][0] = fas;
if (mak1[fas] != u) maxv[u][0] = max(v[u], max1[fas]);
else maxv[u][0] = max(v[u], max2[fas]);
for (int j = 1; j < K; j ++ )
{
fa[u][j] = fa[fa[u][j - 1]][j - 1];
maxv[u][j] = max(maxv[u][j - 1], maxv[fa[u][j - 1]][j - 1]);
}
for (int i = h[u]; i != -1; i = ne[i])
{
int j = e[i];
if (j == fas) continue;
if (mak1[u] != j) sum[j] = max(sum[u], max1[u]) - 2 * w[i];
else sum[j] = max(sum[u], max2[u]) - 2 * w[i];
dfs2(j, u);
}
}
int lca(int a, int b)
{
if (depth[a] < depth[b]) swap(a, b);
for (int j = K - 1; j >= 0; j -- )
if (depth[fa[a][j]] >= depth[b])
{
if (depth[fa[a][j]] == depth[b]) kk = a;
res = max(res, maxv[a][j]);
a = fa[a][j];
}
if (a == b)
{
res = max(res, v[a]);
return a;
}
for (int j = K - 1; j >= 0; j -- )
if (fa[a][j] != fa[b][j])
{
res = max({res, maxv[a][j], maxv[b][j]});
a = fa[a][j];
b = fa[b][j];
}
kk = a;
int u = fa[a][0];
if (a != mak1[u] && b != mak1[u]) res = max(res, max1[u]);
else if (a != mak2[u] && b != mak2[u]) res = max(res, max2[u]);
else res = max(res, max3[u]);
return u;
}
int main()
{
// freopen("speaker4.in", "r", stdin);
// freopen("speaker4.out", "w", stdout);
cin >> n >> m;
for (int i = 1; i <= n; i ++ ) scanf("%lld", &v[i]);
memset(h, -1, sizeof h);
for (int i = 1; i < n; i ++ )
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c);
add(b, a, c);
// cout << a << ' ' << b << ' ' << c << endl;
}
dfs1(1, 0);
dfs2(1, 0);
while (m -- )
{
int a, b;
res = 0;
scanf("%d%d", &a, &b);
int p = lca(a, b);
// printf("%d %d %d\n", a, b, p);
LL ans = max({sum[p], res});
if (p == a)
{
if (mak1[a] != kk) ans = max({ans, max1[b], max1[a]});
else ans = max({ans, max1[b], max2[a]});
}
else if (p == b)
{
if (mak1[b] != kk) ans = max({ans, max1[a], max1[b]});
else ans = max({ans, max1[a], max2[b]});
}
else
{
ans = max({ans, max1[a], max1[b]});
}
printf("%lld\n", ans - (dist[a] + dist[b] - 2 * dist[p]) + v[a] + v[b]);
// printf("%lld %lld %lld %lld %lld\n", sum[p], max1[a], max1[b], res, ans);
}
return 0;
}
</pre><i>参考资料：<a href="https://www.cnblogs.com/blind5883/p/18565823">博客园题解</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P11323">P11323 【MX-S7-T1】「SMOI-R2」Happy Card</a><a href="https://www.luogu.com.cn/problem/list?difficulty=4">普及+/提高</a>这道题需要找规律，构造，分类讨论。最核心的就是将三带一和炸放在一起出。<pre>
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 10;
long long a[MAXN];
int main()
{
int T,n;
long long sum = 0;
long long cnt1 = 0 , cnt2 = 0 , cnt3 = 0;
cin >> T; // 输入询问次数
for (int t = 1 ; t <= T ; ++t)
{
memset(a,0,sizeof(a));
cnt1 = cnt2 = cnt3 = sum = 0;
cin >> n;
for (int i = 1 ; i <= n ; ++i)
{
cin >> a[i]; // 输入每种牌的数量
sum += a[i];
cnt3 += a[i]/3;// 累加可以组成三张的组合
}
if (cnt3 > sum/4) // 如果所有“三张”的总数足以覆盖至少四张一组的分布
{
if (sum % 4 == 1) cout << sum / 4 + 1 << endl;
else if (sum % 4 ==2) cout << sum / 4 + 1 << endl;
else if (sum % 4 == 3) cout << sum / 4 + 2 << endl;
else if (sum % 4 == 0) cout << sum / 4 << endl;
}
else if (cnt3 <= sum/4)
{
for (int i = 1 ; i <= n ; i++)
{
if (a[i] % 3 == 1) cnt1++;// 模 3 余 1 的牌数量
else if (a[i] % 3 == 2) cnt2++;// 模 3 余 2 的牌数量
}
if (cnt3 <= cnt1) cout << cnt1 + cnt2 << endl;// 如果“三张”不足够，使用余数牌
else if (cnt3 > cnt1) cout << cnt3 + cnt2 - (cnt3 - cnt1) / 2 << endl;// 用余数的 2 补充“三带一”或其他牌
}
}
return 0;
}
</pre><i>参考资料：<a href="https://www.luogu.com.cn/problem/solution/P11323">题解区@Exsinly</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P11268">P11268 【MX-S5-T2】买东西题</a><a href="https://www.luogu.com.cn/problem/list?difficulty=4">普及+/提高</a>这道题是反悔背包的题目，不过可以使用另一种解法，也就是构造出n个虚拟优惠券，然后加上题目给出的优惠券，从中挑选n个优惠券最后用不用优惠券的钱来减去全部使用这n个优惠券的钱数就好了。也算是给了我们一个新的思路<pre>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define int long long
struct node{
int x,y;
bool operator < (const node &a)const{
return x>a.x||(x==a.x&&y>a.y);
}
}a[1500000],b[1500000];
int ans,n,m;
signed main(){
cin>>n>>m;
for(int i=1;i<=n;i++){
cin>>a[i].x>>a[i].y;
a[i].y=a[i].x-a[i].y;
ans+=a[i].x;
}
for(int i=1;i<=m;i++){
cin>>b[i].x>>b[i].y;
}
sort(a+1,a+n+1);
sort(b+1,b+m+1);
priority_queue<int,vector<int>,greater<int> > q;
int j=1;
for(int i=1;i<=m;i++){
while(j<=n&&a[j].x>=b[i].x)q.push(a[j].y),j++;
if(!q.empty()&&q.top()<b[i].y){
q.pop();
q.push(b[i].y);
}
}
while(j<=n)q.push(a[j].y),j++;
while(!q.empty()) ans-=q.top(),q.pop();
cout<<ans;
return 0;
}
</pre><i>参考资料：<a href="https://www.cnblogs.com/CodingGoat/p/18536731">博客园题解</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P11267">P11267 【MX-S5-T1】王国边缘</a><a href="https://www.luogu.com.cn/problem/list?difficulty=4">普及+/提高</a>这道题倍增才能满分，不过二分算法可以拿80，为了时间，我还是copy了题解。<pre>
#include<iostream>
#include<cstdio>
#define int long long
using namespace std;
const int mod=1e9+7;
int n,m,q,a[400010],last[400010];
int d1[200010][70],d2[200010][70];
signed main()
{
scanf("%lld%lld%lld",&n,&m,&q);
for(int i=1;i<=n;i++)
{
char c;
cin>>c;
a[i]=a[i+n]=c-'0';
}
for(int i=1;i<=2*n;i++)
{
if(a[i]==1)last[i]=i;
else last[i]=last[i-1];
}
for(int i=1;i<=n;i++)
if(m<n)
{
if(last[i+m]<=i)d1[i][0]=d2[i][0]=1;
else d1[i][0]=d2[i][0]=last[i+m]-i;
}
else
{
if(!last[n])d1[i][0]=1%n,d2[i][0]=1;
else
{
int r=n+(i+m-1)%n+1;
d1[i][0]=(m-r+last[r])%n,d2[i][0]=(m-r+last[r])%mod;
}
}
for(int j=1;j<=61;j++)
for(int i=1;i<=n;i++)
{
int nxt=(i+d1[i][j-1]-1)%n+1;
d1[i][j]=(d1[i][j-1]+d1[nxt][j-1])%n;
d2[i][j]=(d2[i][j-1]+d2[nxt][j-1])%mod;
}
while(q--)
{
int ss,k;
scanf("%lld%lld",&ss,&k);
int now=(ss-1)%n+1,ans=ss%mod;
for(int i=61;i>=0;i--)
if(k>=(1ll<<i))
{
k-=(1ll<<i);
ans=(ans+d2[now][i])%mod;
now=(now+d1[now][i]-1)%n+1;
}
printf("%lld\n",ans);
}
return 0;
}
</pre><i>抄袭资料：<a href="https://www.luogu.com.cn/problem/solution/P11267">题解区@Night_sea_64</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P11290">P11290 【MX-S6-T2】「KDOI-11」飞船</a><a href="https://www.luogu.com.cn/problem/list?difficulty=5">提高+/省选-</a>这道题很难，但是做完却感觉收获满满！首先你能从中学会到dp的高级用法，这是教科书上见不到的。其次，你能对于这种复杂题目更加适应。最后，你还可以学会如何思考这类题目。<pre>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn = 1e5+10;
double f[2][33][22];
double ans[maxn];
pair<int,int> y[maxn];
int n,q;
double t[maxn],s[maxn];
int p[maxn];
int x[maxn];
double pow2[40],pow3[40];
int main(){
pow2[0]=pow3[0]=1;
for(int i=1;i<=32;i++){
pow2[i]=pow2[i-1]*2;
}
for(int i=1;i<=32;i++){
pow3[i]=pow3[i-1]*3;
} <span>//初始化2的n次方和3的n次方</span>
cin>>n>>q;
for(int i=1;i<=n;i++){
cin>>p[i]>>t[i]>>s[i];
} <span>//正常读入数据</span>
for(int i=1;i<=q;i++){
cin>>y[i].first;
y[i].second=i;
} <span>//待会要按照询问端点排序，因此要搞一个pair做离线</span>
for(int i=0;i<2;i++){
for(int j=0;j<33;j++){
for(int k=0;k<23;k++){
f[i][j][k]=1e18;
}
}
} <span>//这就是初始化f数组了，但是我不知道memset怎么对double类型初始化，就干脆写循环了</span>
f[0][0][0]=0; <span>//边界条件你懂的</span>
sort(y+1,y+q+1); <span>//将询问排序</span>
int cury=1; <span>//如字面意思，current就是当前的意思，cury充当了一个目前的指针</span>
for(int i=1;i<=n;i++){
double curp=p[i],befp=p[i-1],ti=t[i];
int xi=s[i];
<span>//这几个变量也都是字面意思，curp目前的p，befp之前的p，ti时间，xi加倍的速度</span>
while(cury<=q&&y[cury].first<curp){ <span>//先处理完所有小于当前加油站的询问</span>
double sum=1e18;
for(int j=0;j<31;j++){
for(int k=0;k<20;k++){
sum=min(sum,f[0][j][k]+(y[cury].first-befp)/pow2[j]/pow3[k]); <span>就是路程除以速度的公式，这里分母是pow2和pow3</span>
<span>//计算结果是时间，相当于预处理将所有可能的时间贡献都枚举出来了</span>
}
} <span>//这就是枚举速度加多少能产生多少时间</span>
ans[y[cury].second]=sum; <span>//选一个最小的作为当前询问的答案</span>
cury++; <span>//指针自增，没问题吧</span>
}
for(int j=31;j>=0;j--){
for(int k=20;k>=0;k--){
f[1][j][k]=f[0][j][k]+(curp-befp)/pow2[j]/pow3[k];
} <span>//前面的while计算的是询问距离befp的时间，这里是当前加油站距离befp的时间</span>
}
for(int j=31;j>=0;j--){
for(int k=20;k>=0;k--){
switch(xi){
case 1:{
break;
}
case 2:{
if(j>0)f[1][j][k]=min(f[1][j][k],f[1][j-1][k]+ti);
break;
}
case 3:{
if(k>0)f[1][j][k]=min(f[1][j][k],f[1][j][k-1]+ti);
break;
}
case 4:{
if(j>1)f[1][j][k]=min(f[1][j][k],f[1][j-2][k]+ti);
break;
}
<span>//根据选项做转移，挺好理解的</span>
}
}
}
for(int j=31;j>=0;j--){
for(int k=20;k>=0;k--){
f[0][j][k]=f[1][j][k];
}
} <span>//我们发现你的时间贡献只跟你上一个加油站有关，因此f第一维只开了2</span>
<span>//因为我们这里是滚动DP，因此要将f[1]的copy给f[0]，就这样滚起来，内存重复利用</span>
}
while(cury<=q){
double sum=1e18;
for(int j=0;j<=31;j++){
for(int k=0;k<=20;k++){
sum=min(sum,f[0][j][k]+(y[cury].first-p[n])/pow2[j]/pow3[k]);
}
}
ans[y[cury].second]=sum;
cury++;
} <span>//这里是处理末尾多出来的尾巴</span>
for(int i=1;i<=q;i++){
printf("%.8lf\n",ans[i]);
} <span>//最后输出答案</span>
return 0;
}
</pre><i>参考资料：<a href="https://www.luogu.com.cn/problem/solution/P11290">题解区@cff_0102</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P11289">P11289 【MX-S6-T1】「KDOI-11」打印</a><a href="https://www.luogu.com.cn/problem/list?difficulty=4">普及+/提高</a>这道题可太恶心了，调了2天才调好。结果居然是因为等待那块没有将now+e[i].s倒直地。<pre>
#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
#define maxn 3000000
#define int long long
struct File
{
int s,t, fileId;
} e[maxn];
bool operator < (File a,File b){
return a.t<b.t;
}
struct Print{
int now,printMachineId;
bool operator < (const Print &a)const{
return now==a.now?printMachineId>a.printMachineId:now>a.now;
}
};
priority_queue<Print> runingPrintMachine,leisurePrintMachine;
priority_queue<int, vector<int>, greater<int>> ans[maxn];
int n, m;
signed main()
{
cin >> n >> m;
for (int i = 1; i <= m; i++)
{
leisurePrintMachine.push({0,i});
}
for (int i = 1, s, t; i <= n; i++)
{
cin >> s >> t;
e[i].s = s, e[i].t = t, e[i].fileId = i;
}
sort(e + 1, e + n + 1);
for (int i = 1; i <= n; i++)
{
while (!runingPrintMachine.empty() && (runingPrintMachine.top().now <= e[i].t))
{
leisurePrintMachine.push({0,runingPrintMachine.top().printMachineId});
runingPrintMachine.pop();
}
int id;
if(!leisurePrintMachine.empty()){
id=leisurePrintMachine.top().printMachineId;
leisurePrintMachine.pop();
runingPrintMachine.push({e[i].t+e[i].s,id});
}
else{
id=runingPrintMachine.top().printMachineId;
runingPrintMachine.push({runingPrintMachine.top().now+e[i].s,id});
runingPrintMachine.pop();
}
ans[id].push(e[i].fileId);
}
for (int i = 1; i <= m; i++)
{
cout << ans[i].size() << " ";
while (!ans[i].empty())
{
cout << ans[i].top() << " ";
ans[i].pop();
}
cout << '\n';
}
return 0;
}
</pre><i>参考资料：<a href="https://www.luogu.com.cn/problem/solution/P11289">题解区@zhujiangyuan</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P9871">P9871 [NOIP2023] 天天爱打卡</a><a href="https://www.luogu.com.cn/problem/list?difficulty=6">省选/NOI-</a>这道题还是根据出题人给的数据范围和特殊性质一步一步引导你想出正解，但是我却想不出来，因此我甚至直接copy。<pre>
#include<bits/stdc++.h>
#define ll long long
#define mid (l+r>>1)
#define rep(x,y,z) for(int x=(y);x<=(z);x++)
#define per(x,y,z) for(int x=(y);x>=(z);x--)
#define e(x) for(int i=h[x],y=to[i];i;i=nxt[i],y=to[i])
#define Pi pair<int,int>
#define pb push_back
#define ui unsigned ll
inline int read(){int s=0,w=1;char c=getchar();while(c<48||c>57){if(c=='-')w=-1;c=getchar();}while(c>=48&&c<=57)s=(s<<1)+(s<<3)+c-'0',c=getchar();return s*w;}
const int N=2e5+5,M=1e6+5,inf=(1LL<<31)-1;
const ll llf=1e18;
using namespace std;
int n,m,k,d,lsh[N],b[N],ln,testid,T;//days,tasks,limits,costs
struct node{
int l,r,k;
}a[N];
inline void init(){
ln=0;
n=read(),m=read(),k=read(),d=read();
lsh[m*2+1]=0;
rep(i,1,m){
a[i].r=read(),a[i].l=a[i].r-read()+1,a[i].k=read();
lsh[(i<<1)-1]=a[i].l,lsh[i<<1]=a[i].r;
}
sort(lsh+1,lsh+m*2+1);
rep(i,2,m*2+1)if(lsh[i]^lsh[i-1])b[++ln]=lsh[i-1];//离散化
}
ll dp[N];
inline ll got(int x){
if(x>=0)return dp[x];
return 0;
}
vector<Pi >p[N];
#define E(x) for(auto y:p[x])
#define L x<<1
#define R L|1
#define lc L,l,mid
#define rc R,mid+1,r
#define OK Ll<=l&&r<=Rr
struct seg{
ll w,laz;
}xd[N<<2];
inline void insert(int x,ll k){
xd[x].w+=k,xd[x].laz+=k;
}
inline void pushdown(int x){
insert(L,xd[x].laz),insert(R,xd[x].laz),xd[x].laz=0;
}
inline void getup(int x){
xd[x].w=max(xd[L].w,xd[R].w);
}
inline void build(int x,int l,int r){
xd[x]={0,0};
if(l==r)return;
build(lc),build(rc);
}
inline void modify(int x,int l,int r,int Ll,int Rr,ll k){
if(OK)return insert(x,k),void();
pushdown(x);
if(Ll<=mid&&Rr>=l)modify(lc,Ll,Rr,k);
if(Ll<=r&&Rr>mid)modify(rc,Ll,Rr,k);
getup(x);
}
inline ll query(int x,int l,int r,int Ll,int Rr){
if(Ll>Rr)return 0;
if(OK)return xd[x].w;
pushdown(x);
if(Rr<=mid)return query(lc,Ll,Rr);
if(Ll>mid)return query(rc,Ll,Rr);
return max(query(lc,Ll,Rr),query(rc,Ll,Rr));
}
//线段树
#define Root 1,1,ln
inline int find(int x){
int l=1,r=ln,ans=r;
while(l<=r)if(b[mid]>=x)ans=mid,r=mid-1;
else l=mid+1;
return ans;
}//手写二分
inline void subtaskall(){
build(Root);
rep(i,0,ln)dp[i]=0;
rep(i,1,m){
a[i].l=lower_bound(b+1,b+ln+1,a[i].l)-b;
a[i].r=lower_bound(b+1,b+ln+1,a[i].r)-b;
p[a[i].r].pb({a[i].l,a[i].k});
}
rep(i,1,ln){
E(i)modify(Root,1,y.first,y.second);
int j=find(b[i]-k+1);
dp[i]=max(dp[i],query(Root,j,i-1))-1LL*b[i]*d-1LL*d;
int pos=i-2;
if(b[i-1]!=b[i]-1)pos=i-1;
dp[i]=max(dp[i],query(Root,i,i)+got(pos)-d);
dp[i]=max(dp[i],dp[i-1]);
modify(Root,i,i,got(pos)+1LL*b[i]*d);
}
cout <<dp[ln]<<'\n';
rep(i,1,ln)p[i].clear();
}//dp
inline void solve(){
subtaskall();
}
int main(){
testid=read(),T=read();
while(T--){
init();
solve();
}
return 0;
}
</pre><i>参考资料：<a href="https://www.luogu.com.cn/problem/solution/P9871">题解区@未来姚班zyl</a></i></div>
<div class="t1"><a href="https://www.luogu.com.cn/problem/P9870">P9870 [NOIP2023] 双序列拓展</a><a href="https://www.luogu.com.cn/problem/list?difficulty=6">省选/NOI-</a>这道题是可以根据性质推出解决方案的。35分做法：<b>如果f[i]<=g[j]，那么我们就拓展f[i-1]直到f[i]>g[j]。换句话说就是这个：<img alt="" src="https://upload-bbs.miyoushe.com/upload/2024/11/26/198629752/5c64e8c2798d2b864691978c7077c87b_8578966308029315410.png"/></b><pre>
#include <iostream>
#include <cstdio>
#include <cstring>