网络流之最大流（增广路）

hdu3459

Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 27520    Accepted Submission(s): 12155   Problem Description Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.     Input The 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)     Output For 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

最大流模板题

 

#include
using namespace std;
#define INF 0x3f3f3f3f
int t,ca,cap[20][20],flow[20][20],n,m,a[20],p[20];
//cap 容量
//flow 流量
//a 增广路的最大流量 
int main()
{
	scanf("%d",&t);
	ca=1;
	while(t--)
	{
		memset(cap,0,sizeof(cap));
		memset(flow,0,sizeof(flow));//初始化
		scanf("%d%d",&n,&m);
		for(int i=1; i<=m; i++)
		{
			int x,y,z;
			scanf("%d%d%d",&x,&y,&z);
			cap[x][y]+=z;//x到y的容量增加z 
		}
		int s=1,t=n;//起点为 1  终点为 n 
		int f=0;//最大流的初始值为 0 
		queueq;
		while(1)
		{
			memset(a,0,sizeof(a));
			a[s]=INF;
			q.push(s);
			while(!q.empty())
			{
				int u=q.front();
				q.pop();
				for(int i=1; i<=n; i++)
				{
					if(a[i]==0&&cap[u][i]>flow[u][i])//如果这个点没被走过     并且容量大于现在容量(可以产生增广路) 
					{
						p[i]=u;//让这个点的爸爸是 u 表示 i 是从 u 走过来的  
						q.push(i);//入队 
						a[i]=min(a[u],cap[u][i]-flow[u][i]);
						//上一个点流过来的量为 a[u] 这个点能通过的最大流量为 cap[u][i]-flow[u][i] 
						//去俩者的最小值 为这个点增广路的流量 
					}
				}
			}
			if(a[t]==0) break;//如果这条路没有到终点 那么代表已经不能构成增广路了 
			for(int u=t; u!=s; u=p[u])//从终点开始 往前找到这条增广路的每个点 
			{
				flow[p[u]][u]+=a[t];//正流量加 
				flow[u][p[u]]-=a[t];//负流量减 
			}
			f+=a[t];//总流量加上这条增广路最后的流量 
		}
		printf("Case %d: %d\n", ca++, f);
	}
	return 0;
}