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  • hdoj 3549 Flow Problem【网络流最大流入门】

    Flow Problem

    Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
    Total Submission(s): 11405    Accepted Submission(s): 5418


    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
     
     
    刚开始看不是太理解  解析会后续更新
    #include<stdio.h>
    #include<string.h>
    #include<stack>
    #include<queue>
    #include<vector>
    #include<algorithm>
    #define INF 0x7fffff
    #define MAX 2100
    using namespace std;
    int ans,head[MAX];
    int n,m;
    int dis[MAX],vis[MAX];
    int cur[MAX];
    struct node
    {
    	int beg,end,cap,flow,next;
    }edge[MAX];
    void init()
    {
    	ans=0;
    	memset(head,-1,sizeof(head));
    }
    void add(int u,int v,int w)
    {
    	edge[ans].beg=u;
    	edge[ans].end=v;
    	edge[ans].cap=w;
    	edge[ans].flow=0;
    	edge[ans].next=head[u];
    	head[u]=ans++;
    }
    void getmap()
    {
    	int i,a,b,c;
    	while(m--)
    	{
    		scanf("%d%d%d",&a,&b,&c);
    		add(a,b,c);
    		add(b,a,0);
    	}
    }
    int bfs(int start,int over)
    {
    	int i,v;
    	memset(dis,-1,sizeof(dis));
    	memset(vis,0,sizeof(vis));
        queue<int>q;
        while(!q.empty())
            q.pop();
        q.push(start);
        vis[start]=1;
        dis[start]=0;
        while(!q.empty())
        {
        	int u=q.front();
        	q.pop();
        	for(i=head[u];i!=-1;i=edge[i].next)
        	{
        	    v=edge[i].end;
    			if(!vis[v]&&edge[i].cap>edge[i].flow)
    			{
    				vis[v]=1;
    				dis[v]=dis[u]+1;
    				if(v==over)
    				    return 1;
    				q.push(v);
    			}	
        	}
        }
        return 0;
    }
    int dfs(int x,int a,int over)
    {
    	if(x==over||a==0)
    	    return a;
    	int flow=0,f;
    	for(int& i=cur[x];i!=-1;i=edge[i].next)
    	{
    		if(dis[x]+1==dis[edge[i].end]&&(f=dfs(edge[i].end,min(a,edge[i].cap-edge[i].flow),over))>0)
    		{
    			edge[i].flow+=f;
    			edge[i^1].flow-=f;
    			flow+=f;
    			a-=f;
    			if(a==0) break;
    		}
    	}
    	return flow;
    }
    int maxflow(int start,int over)
    {
    	int flow=0;
    	while(bfs(start,over))
    	{
    		memcpy(cur,head,sizeof(head));
    		flow+=dfs(start,INF,over);
    	}
    	return flow;
    }
    int main()
    {
    	int t,k,j,i;
    	scanf("%d",&t);
    	k=1;
    	while(t--)
    	{
    		scanf("%d%d",&n,&m);
    		init();
    		getmap();
    		printf("Case %d: ",k++);
    		printf("%d
    ",maxflow(1,n));
    	}
    	return 0;
    }
    

      

     
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  • 原文地址:https://www.cnblogs.com/tonghao/p/4906467.html
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