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  • hdoj--3549--Flow Problem(最大流)

    Flow Problem

    Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
    Total Submission(s): 11733    Accepted Submission(s): 5565



    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
     

    Author
    HyperHexagon
     

    Source
     

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    #include<stdio.h>
    #include<string.h>
    #include<queue>
    #include<vector>
    #include<algorithm>
    using namespace std;
    #define MAX 200+10
    #define INF 10000000+10
    struct node
    {
    	int u,v,cap,flow,next;
    }edge[2100];
    int head[MAX],cur[MAX];
    int vis[MAX],dis[MAX],m,n,top;
    void init()
    {
    	top=0;
    	memset(head,-1,sizeof(head));
    }
    void add(int a,int b,int c)
    {
    	node E1={a,b,c,0,head[a]};
    	edge[top]=E1;
    	head[a]=top++;
    	node E2={b,a,0,0,head[b]};
    	edge[top]=E2;
    	head[b]=top++;
    }
    bool bfs(int s,int e)
    {
    	memset(vis,0,sizeof(vis));
    	memset(dis,-1,sizeof(dis));
    	queue<int>q;
    	while(!q.empty()) q.pop();
    	q.push(s);
    	vis[s]=1;
    	dis[s]=0;
    	while(!q.empty())
    	{
    		int u=q.front();
    		q.pop();
    		for(int i=head[u];i!=-1;i=edge[i].next)
    		{
    			node E=edge[i];
    			if(!vis[E.v]&&E.cap>E.flow)
    			{
    				vis[E.v]=1;
    				dis[E.v]=dis[E.u]+1;
    				if(E.v==e)
    				return true;
    				q.push(E.v);
    			}
    		}
    	}
    	return false;
    }
    int dfs(int x,int a,int e)
    {
    	if(x==e||a==0) return a;
    	int flow=0,f;
    	for(int i=cur[x];i!=-1;i=edge[i].next)
    	{
    		node& E=edge[i];
    		if(dis[x]+1==dis[E.v]&&(f=dfs(E.v,min(a,E.cap-E.flow),e))>0)
    		{
    			E.flow+=f;
    			flow+=f;
    			edge[i^1].flow-=f;
    			a-=f;
    			if(a==0) break;
    		}
    	}
    	return flow;
    }
    int MAXflow(int s,int e)
    {
    	int flow=0;
    	while(bfs(s,e))
    	{
    		memcpy(cur,head,sizeof(head));
    		flow+=dfs(s,INF,e);
    	}
    	return flow;
    }
    void getmap()
    {
    	int a,b,c;
    	while(m--)
    	{
    		scanf("%d%d%d",&a,&b,&c);
    		add(a,b,c);
    	}
    }
    int main()
    {
    	int t;
    	int k=1;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%d%d",&n,&m);
    		init();
    		getmap();
    		printf("Case %d: %d
    ",k++,MAXflow(1,n));
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273652.html
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