hdu 6214 Smallest Minimum Cut[最大流]

Posted 2020-10-07 GraceSkyer

hdu 6214 Smallest Minimum Cut[最大流]

题意：求最小割中最少的边数。

题解：对边权乘个比边大点的数比如300，再加1 ，最后，最大流对300取余就是边数啦。。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<vector>
 7 #include<queue>
 8 using namespace std;
 9 #define CLR(a,b) memset((a),(b),sizeof((a)))
10 typedef long long ll;
11 const int N = 205;
12 const int inf = 0x3f3f3f3f;
13 int n, m, S, T;
14 int dep[N], cur[N];
15 int head[N];
16 struct Edge{
17     int v, c, nex;
18     Edge(int _v=0,int _c=0,int _nex=0):v(_v),c(_c),nex(_nex){}
19 };
20 vector<Edge>E;
21 void add(int u,int v,int c){E.push_back(Edge(v,c,head[u]));head[u]=E.size()-1;}
22 
23 bool bfs() {
24     queue<int> q;
25     CLR(dep, -1);
26     q.push(S); dep[S] = 0;
27     while(!q.empty()) {
28         int u = q.front(); q.pop();
29         for(int i = head[u]; ~i; i = E[i].nex) {
30             int v = E[i].v;
31             if(E[i].c && dep[v] == -1) {
32                 dep[v] = dep[u] + 1;
33                 q.push(v);
34             }
35         }
36     }
37     return dep[T] != -1;
38 }
39 int dfs(int u, int flow) {
40     if(u == T) return flow;
41     int w, used=0;
42     for(int i = head[u]; ~i; i = E[i].nex) {
43         int v = E[i].v;
44         if(dep[v] == dep[u] + 1) {
45             w = flow - used;
46             w = dfs(v, min(w, E[i].c));
47             E[i].c -= w;  E[i^1].c += w;
48             if(v) cur[u] = i;
49             used += w;
50             if(used == flow) return flow;
51         }
52     }
53     if(!used) dep[u] = -1;
54     return used;
55 }
56 int dinic() {
57     int ans = 0;
58     while(bfs()) {
59         for(int i = 1; i <= T;i++)
60             cur[i] = head[i];
61         ans += dfs(S, inf);
62     }
63     return ans;
64 }
65 int main() {
66     int t, i, u, v, w;
67     scanf("%d", &t);
68     while(t--) {
69         E.clear(); CLR(head, -1);
70         scanf("%d%d", &n, &m);
71         scanf("%d%d", &S, &T);
72         for(i = 1; i <= m; ++i) {
73             scanf("%d%d%d", &u, &v, &w);
74             add(u, v, w*300+1); add(v, u, 0);
75         }
76         int ans = dinic()%300;
77         printf("%d\n", ans);
78     }
79     return 0;
80 }

249ms