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 ", ans); 78 } 79 return 0; 80 }