zoukankan      html  css  js  c++  java
  • Constructing Roads*

    Constructing Roads

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 10700    Accepted Submission(s): 3989
    Problem Description
    There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected. 

    We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
     

    Input
    The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

    Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
     

    Output
    You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum. 
     

    Sample Input
    3 0 990 692 990 0 179 692 179 0 1 1 2
     

    Sample Output
    179
     
    #include <iostream>
    using namespace std;
    const int INF=0x3f3f3f3f;
    const int MAX=105;
    int cost[MAX][MAX];
    int lowcost[MAX];
    void prim(int n)
    {
    	int i,j;
    	int sum=0;
    	for(i=2;i<=n;i++)
    	{
    		lowcost[i]=cost[1][i];
    	}
    	int min,close;
    	for(i=2;i<=n;i++)//**从2开始吧。1已经选进去了。
    	{
    		min=lowcost[i];
    		close=i;
    		for(j=2;j<=n;j++)
    			if(lowcost[j]<min)
    			{
    				min=lowcost[j];
    				close=j;
    			}
    				sum+=min;
    				lowcost[close]=INF;//**原来错误的做法赋值为0
    				for(j=2;j<=n;j++)
    					if(cost[close][j]<lowcost[j]&&lowcost[j]<INF)//**然后这里用不等于0来判断,其实不对的。
    					{
    						lowcost[j]=cost[close][j];
    					}
    	}
    	printf("%d
    ",sum);
    }
    int main()
    {
    	int n,q;
    	while(scanf("%d",&n)!=EOF)
    	{
    		int i,j;
    		for(i=1;i<=n;i++)
    			for(j=1;j<=n;j++)
    			{
    				scanf("%d",&cost[i][j]);
    			}
    		scanf("%d",&q);
    		int a,b;
    		while(q--)
    		{
    			scanf("%d%d",&a,&b);
    			cost[a][b]=cost[b][a]=0;
    		}
    		prim(n);
    	}
    	return 0;
    }
    /*这题是个好prim算法的模板*/


  • 相关阅读:
    leetcode 189. Rotate Array 数组旋转 ---------- java
    Google test Problem A. Country Leader
    leetcode 187. Repeated DNA Sequences 求重复的DNA串 ---------- java
    mysql忘记密码(未初始化)
    leetcode 186. Reverse Words in a String II 旋转字符数组 ---------- java
    CSS3属性transform详解之(旋转:rotate,缩放:scale,倾斜:skew,移动:translate)(转载)
    bootstrap
    bootstrap使用中遇到的问题(二)
    兼容ie8 rgba()用法
    浏览器前缀
  • 原文地址:https://www.cnblogs.com/unclejelly/p/4082147.html
Copyright © 2011-2022 走看看