标签：can ase min inf net main rect rom next
Flow Problem HDU - 3549
InputThe first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)OutputFor each test cases, you should output the maximum flow from source 1 to sink N.Sample Input
2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1
Sample Output
Case 1: 1 Case 2: 2
题意：网络流模版题，1到n的结点的最大的流
思路：用dinic可以过，为什么之前用sap的邻接矩阵和邻接表都wa或者T了呢
#include<stdio.h> #include<iostream> #include<map> #include<string.h> #include<queue> #include<vector> #include<math.h> #include<algorithm> #define inf 0x3f3f3f3f using namespace std; int dis[20]; int flow[500][500]; int n,m; int bfs() { memset(dis,-1,sizeof(dis)); queue<int>Q; dis[1]=1; Q.push(1); while(!Q.empty()) { int top=Q.front(); Q.pop(); for(int i=1;i<=n;i++) { if(flow[top][i]>0&&dis[i]<0) { dis[i]=dis[top]+1; Q.push(i); } } } if(dis[n]>0)return 1; return 0; } int dinic(int x,int k) { if(x==n) return k; int y; for(int i=1;i<=n;i++) { if(flow[x][i]>0&&dis[i]==dis[x]+1&&(y=dinic(i,min(k,flow[x][i])))) { flow[x][i]-=y; flow[i][x]+=y; return y; } } return 0; } int main() { int t; scanf("%d",&t); int cas=1; while(t--) { memset(flow,0,sizeof(flow)); scanf("%d%d",&n,&m); int a,b,c; for(int i=0;i<m;i++) { scanf("%d%d%d",&a,&b,&c); flow[a][b]+=c; } int ans=0; while(bfs()) { //cout<<"++"<<endl; int res; while(res=dinic(1,inf))ans+=res; } printf("Case %d: %d\n",cas++,ans); } }
标签：can ase min inf net main rect rom next
原文地址：https://www.cnblogs.com/smallhester/p/10295624.html
