zoukankan      html  css  js  c++  java
  • hdoj 3371 Connect the Cities

    Connect the Cities

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 12903    Accepted Submission(s): 3549


    Problem Description
    In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.  
     
    Input
    The first line contains the number of test cases.
    Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
    To make it easy, the cities are signed from 1 to n.
    Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
    Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.
     
    Output
    For each case, output the least money you need to take, if it’s impossible, just output -1.
     
    Sample Input
    1
    6 4 3
    1 4 2
    2 6 1
    2 3 5
    3 4 33
    2 1 2
    2 1 3
    3 4 5 6
     
    Sample Output
    1
     
    prime算法
    #include<stdio.h>
    #include<string.h>
    #define MAX 550
    #define INF 0x3f3f3f
    int city;
    int visit[MAX],map[MAX][MAX],low[MAX];
    void prime()
    {
    	int j,i,min,mincost=0,next;
    	memset(visit,0,sizeof(visit));
    	for(i=1;i<=city;i++)
    	{
    		low[i]=map[1][i];
    	}
    	visit[1]=1;
    	for(i=1;i<city;i++)
    	{
    		min=INF;
    		for(j=1;j<=city;j++)
    		{
    			if(!visit[j]&&min>low[j])
    			{
    				next=j;
    				min=low[j];
    			}
    		}
    		if(min==INF)
    		{
    			printf("-1
    ");
    			return ;
    		}
    		mincost+=min;
    		visit[next]=1;
    		for(j=1;j<=city;j++)
    		{
    			if(!visit[j]&&low[j]>map[next][j])
    			low[j]=map[next][j];
    		}
    	}
    	printf("%d
    ",mincost);
    }
    int main()
    {
    	int n,m,j,i,k,q,p,c,t,x,y,l;
    	int a[MAX];
    	scanf("%d",&n);
    	while(n--)
    	{
    		scanf("%d%d%d",&city,&m,&k);
    		for(i=1;i<=city;i++)
    		{
    			for(j=1;j<=city;j++)
    			{
    				if(i==j)
    				map[i][j]=0;
    				else
    				map[i][j]=INF;
    			}
    		}	
    		for(i=1;i<=m;i++)
    		{
    			scanf("%d%d%d",&p,&q,&c);
    			if(map[p][q]>c)
    			map[p][q]=map[q][p]=c;
    		}
    		while(k--)
    		{			
    			scanf("%d",&t);
    			for(l=1;l<=t;l++)
    			scanf("%d",&a[l]);
    			for(i=1;i<t;i++)
    			{
    				for(j=i+1;j<=t;j++)
    				{
    					map[a[i]][a[j]]=map[a[j]][a[i]]=0;
    				}
    			}
    		}
    		prime();
    	}
    	return 0;
    }
    

      

  • 相关阅读:
    jQuery选择器汇总
    jQuery源码分析系列:总体架构
    jQuery源码分析系列:队列操作
    jQuery源码分析系列:事件
    jQuery源码分析系列:数据缓存
    jQuery源码分析系列:Deferred延迟队列
    Redis基本数据类型与内部存储结构
    oracle 学习笔记1
    设计模式学习1
    注册DEV控件
  • 原文地址:https://www.cnblogs.com/tonghao/p/4542415.html
Copyright © 2011-2022 走看看