-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathmin_cost.cpp
More file actions
139 lines (131 loc) · 6.01 KB
/
min_cost.cpp
File metadata and controls
139 lines (131 loc) · 6.01 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pb push_back
#define mp make_pair
#define all(a) begin(a),end(a)
#define FOR(x,val,to) for(int x=(val);x<int((to));++x)
#define FORE(x,val,to) for(auto x=(val);x<=(to);++x)
#define FORR(x,arr) for(auto &x: arr)
#define FORS(x,plus,arr) for(auto x = begin(arr)+(plus); x != end(arr); ++x)
#define FORREV(x,plus,arr) for(auto x = (arr).rbegin()+(plus); x !=(arr).rend(); ++x)
#define REE(s_) {cout<<s_<<'\n';exit(0);}
#define GET(arr) for(auto &i: (arr)) sc(i)
#define whatis(x) cerr << #x << " is " << (x) << endl;
#define e1 first
#define e2 second
#define INF 0x7f7f7f7f
typedef std::pair<int,int> pi;
typedef std::vector<int> vi;
typedef std::vector<std::string> vs;
typedef int64_t ll;
typedef uint64_t ull;
#define umap unordered_map
#define uset unordered_set
using namespace std;
using namespace __gnu_pbds;
#ifdef ONLINE_JUDGE
#define whatis(x) ;
#endif
#ifdef _WIN32
#define getchar_unlocked() _getchar_nolock()
#define _CRT_DISABLE_PERFCRIT_LOCKS
#endif
template<class L, class R> ostream& operator<<(ostream &os, map<L, R> P) { for(auto const &vv: P)os<<"("<<vv.first<<","<<vv.second<<")"; return os; }
template<class T> ostream& operator<<(ostream &os, set<T> V) { os<<"[";for(auto const &vv:V)os<<vv<<","; os<<"]"; return os; }
template<class T> ostream& operator<<(ostream &os, vector<T> V) { os<<"[";for(auto const &vv:V)os<<vv<<","; os<<"]"; return os; }
template<class L, class R> ostream& operator<<(ostream &os, pair<L, R> P) { os<<"("<<P.first<<","<<P.second<<")"; return os; }
inline int fstoi(const string &str){auto it=str.begin();bool neg=0;int num=0;if(*it=='-')neg=1;else num=*it-'0';++it;while(it<str.end()) num=num*10+(*it++-'0');if(neg)num*=-1;return num;}
inline void getch(char &x){while(x = getchar_unlocked(), x < 33){;}}
inline void getstr(string &str){str.clear(); char cur;while(cur=getchar_unlocked(),cur<33){;}while(cur>32){str+=cur;cur=getchar_unlocked();}}
template<typename T> inline bool sc(T &num){ bool neg=0; int c; num=0; while(c=getchar_unlocked(),c<33){if(c == EOF) return false;} if(c=='-'){ neg=1; c=getchar_unlocked(); } for(;c>47;c=getchar_unlocked()) num=num*10+c-48; if(neg) num*=-1; return true;}template<typename T, typename ...Args> inline void sc(T &num, Args &...args){ bool neg=0; int c; num=0; while(c=getchar_unlocked(),c<33){;} if(c=='-'){ neg=1; c=getchar_unlocked(); } for(;c>47;c=getchar_unlocked()) num=num*10+c-48; if(neg) num*=-1; sc(args...); }
template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; //s.find_by_order(), s.order_of_key() <- works like lower_bound
template<typename T> using ordered_map = tree<T, int, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// Upewnij się, ile dokładnie każdy node może (maksymalnie) dostać flowu (Druz1B flashbacks)
// Czasami, jeśli dany node ma wiele ograniczonych in-edgy, a ma mieć
// ograniczone out-edgea, to trzeba dodać dodatkowy node z jednym inem od tego,
// aby na pewno ograniczenia co do całkowitego out-flowu z tego nodea się zgadzały
// Btw, ten graf nie może mieć ujemnych cykli -> Przed aplikowaniem mcmf trzeba
// tak przekształcić graf, aby nie miał ujemnych cykli, czyli either jakaś
// konwersja do scc może być potrzebna, albo findowanie edgea z najmniejszym
// capacity w cyklu, i zastąpianie cyklu jednym nodem będącym bottleneckiem
// jakoś (https://codeforces.com/blog/entry/53404).
// Also, wtf, this implementation had TLE on atcoder problem with 4s tle, while
// atcoder library mcmf passed in 257ms (though it doesn't support negative
// edges). This version didn't pass even without negative edges.
// https://atcoder.jp/contests/abc214/submissions/25128817 vs https://atcoder.jp/contests/abc214/submissions/25128832
// Tomasz Nowak, Michał Staniewski
struct MCMF {
struct Edge {
int v, u, flow, cap;
ll cost;
friend ostream& operator<<(ostream &os, Edge &e) {
return os << vector<ll>{e.v, e.u, e.flow, e.cap, e.cost};
}
};
int n;
const ll inf_ll = 1e18;
const int inf_int = 1e9;
vector<vector<int>> graph;
vector<Edge> edges;
MCMF(int N) : n(N), graph(n) {}
void add_edge(int v, int u, int cap, ll cost) {
int e = edges.size();
graph[v].emplace_back(e);
graph[u].emplace_back(e + 1);
edges.emplace_back(Edge{v, u, 0, cap, cost});
edges.emplace_back(Edge{u, v, 0, 0, -cost});
}
pair<int, ll> augment(int source, int sink) {
vector<ll> dist(n, inf_ll);
vector<int> from(n, -1);
dist[source] = 0;
deque<int> que = {source};
vector<bool> inside(n);
inside[source] = true;
while(que.size()) {
int v = que.front();
inside[v] = false;
que.pop_front();
for(int i : graph[v]) {
Edge &e = edges[i];
if(e.flow != e.cap and dist[e.u] > dist[v] + e.cost) {
dist[e.u] = dist[v] + e.cost;
from[e.u] = i;
if(not inside[e.u]) {
inside[e.u] = true;
que.emplace_back(e.u);
}
}
}
}
if(from[sink] == -1)
return {0, 0};
int flow = inf_int, e = from[sink];
while(e != -1) {
flow = min(flow, edges[e].cap - edges[e].flow);
e = from[edges[e].v];
}
e = from[sink];
while(e != -1){
edges[e].flow += flow;
edges[e ^ 1].flow -= flow;
e = from[edges[e].v];
}
return {flow, flow * dist[sink]};
}
pair<int, ll> operator()(int source, int sink) {
int flow = 0;
ll cost = 0;
pair<int, ll> got;
do {
got = augment(source, sink);
flow += got.first;
cost += got.second;
} while(got.first);
return {flow, cost};
}
};
int main(){
ios_base::sync_with_stdio(0);cin.tie(0);
}