zoukankan      html  css  js  c++  java
  • hdoj--3488--Tour(KM)

    Tour

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
    Total Submission(s): 2549    Accepted Submission(s): 1257


    Problem Description
    In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000) one-way roads connecting them. You are lucky enough to have a chance to have a tour in the kingdom. The route should be designed as: The route should contain one or more loops. (A loop is a route like: A->B->……->P->A.)
    Every city should be just in one route.
    A loop should have at least two cities. In one route, each city should be visited just once. (The only exception is that the first and the last city should be the same and this city is visited twice.)
    The total distance the N roads you have chosen should be minimized.
     

    Input
    An integer T in the first line indicates the number of the test cases.
    In each test case, the first line contains two integers N and M, indicating the number of the cities and the one-way roads. Then M lines followed, each line has three integers U, V and W (0 < W <= 10000), indicating that there is a road from U to V, with the distance of W.
    It is guaranteed that at least one valid arrangement of the tour is existed.
    A blank line is followed after each test case.
     

    Output
    For each test case, output a line with exactly one integer, which is the minimum total distance.
     

    Sample Input
    1 6 9 1 2 5 2 3 5 3 1 10 3 4 12 4 1 8 4 6 11 5 4 7 5 6 9 6 5 4
     

    Sample Output
    42
     

    Source
     

    Recommend
    zhouzeyong   |   We have carefully selected several similar problems for you:  3435 1853 3395 3491 3657 

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #define INF 0x3f3f3f3f
    using namespace std;
    int n,m;
    int map[210][210];
    int slock[210];
    int lx[210],ly[210];
    bool visx[210],visy[210];
    int nx,ny,match[210];
    void getmap()
    {
    	scanf("%d%d",&n,&m);
    	nx=ny=n;
    	for(int i=1;i<=nx;i++)
    	for(int j=1;j<=ny;j++)
    	map[i][j]=-INF;
    	int x,y,w;
    	for(int i=0;i<m;i++)
    	{
    		scanf("%d%d%d",&x,&y,&w);
    		if(-w>map[x][y])
    			map[x][y]=-w;
    	}
    }
    int DFS(int x)
    {
    	visx[x]=true;
    	for(int y=1;y<=ny;y++)
    	{
    		if(visy[y]) continue;
    		int t=lx[x]+ly[y]-map[x][y];
    		if(t==0)
    		{
    			visy[y]=true;
    			if(match[y]==-1||DFS(match[y]))
    			{
    				match[y]=x;
    				return 1;
    			}
    		}
    		else if(t<slock[y])
    		slock[y]=t;
    	}
    	return 0;
    }
    void KM()
    {
    	memset(match,-1,sizeof(match));
    	memset(ly,0,sizeof(ly));
    	for(int x=1;x<=nx;x++)
    	{
    		lx[x]=-INF;
    		for(int y=1;y<=ny;y++)
    		lx[x]=max(lx[x],map[x][y]);
    	}
    	for(int x=1;x<=nx;x++)
    	{
    		for(int y=1;y<=ny;y++)
    		slock[y]=INF;
    		while(1)
    		{
    			memset(visx,false,sizeof(visx));
    			memset(visy,false,sizeof(visy));
    			if(DFS(x)) break;
    			int d=INF;
    			for(int i=1;i<=ny;i++)
    			{
    				if(!visy[i]&&slock[i]<d)
    					d=slock[i];
    			}
    			for(int i=1;i<=nx;i++)
    			{
    				if(visx[i])
    				lx[i]-=d;
    			}
    			for(int i=1;i<=ny;i++)
    			{
    				if(visy[i])
    					ly[i]+=d;
    				else
    					slock[i]-=d;
    			}
    		}
    	}
    	int ans = 0;
    	for(int i = 1;i <= ny; i++)
    	ans += map[match[i]][i];
    	printf("%d
    ",-ans); 
    }
    int main()
    {
    	int t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		getmap();
    		KM();
    	}
    	return 0;
    }


  • 相关阅读:
    touchMove VS touchCancel
    svg viewbox 作用
    reactjs reactLink
    放开linux下的端口
    运算符重载函数作为类成员函数和友元函数 (转)
    MBean和MXBean 区别
    transfer-encoding
    CSRF
    vue知识拓展
    居中
  • 原文地址:https://www.cnblogs.com/playboy307/p/5273477.html
Copyright © 2011-2022 走看看