zoukankan      html  css  js  c++  java
  • 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
     

    Recommend
    zhengfeng   |   We have carefully selected several similar problems for you:  1532 3572 3416 3081 3491

    数组大小也是一门学问啊


    #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;
    }


  • 相关阅读:
    js中用setTimeout写定时炸弹
    javascript函数的形参和实参
    在Mac OS X中配置Apache+PHP+MySQL(转)
    600多万用户密码使用次数统计分析程序
    pgsql数据库创建以及压缩包形式导入导出
    apache中的.htaccess配置示例收集整理
    收集常用的正则表达式及其应用
    采用htpasswd设置验证机制
    用apache做代理时候,可设置请求头信息
    ubuntu在混合环境中创建和配置文件共享
  • 原文地址:https://www.cnblogs.com/playboy307/p/5273652.html
Copyright © 2011-2022 走看看