Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

Source

kicc

还是修路问题。这道题两种解法，一种是贪心+并查集。

一种就是最小生成树了。今天刚学的。

所以用了它做。发现自己无法用语言来描绘这个算法。。

贴下代码吧。

#include <stdio.h>
#include <algorithm>
using namespace std;
#include <string.h>
#define inf 0x6ffffff
int vis[102];//用来标记是否被用过
int map[102][102];
int dis[102];
int n,sum;
void prim()
{
	int i,j,k,min;
	for(i=1;i<=n;i++)
		dis[i]=map[1][i];
	vis[1]=1;
	   for(i=1;i<=n;i++)
	   {
		   min=inf;
		   for(j=1;j<=n;j++)
		   {
			   if(vis[j]==0&&dis[j]<min)
			   {
				   min=dis[j];
				   k=j;
			   }
		   }
		   if(min==inf)
			   break;
		   vis[k]=1;
		   sum+=min;
		   for(j=1;j<=n;j++)
		   {
			   if(vis[j]==0 &&dis[j]>map[k][j])
				   dis[j]=map[k][j];
		   }
	   }
}
int main()
{
	int ll,i,j,str,end,m;
	while(scanf("%d",&n)!=EOF)
	{
		memset(vis,0,sizeof(vis));
		memset(dis,0,sizeof(dis));
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
				map[i][j]=inf;
			for(i=1;i<=n;i++)
				for(j=1;j<=n;j++)
				{
					scanf("%d",&ll);
					if(map[i][j]>ll)
						map[i][j]=map[j][i]=ll;
				}
				scanf("%d",&m);
				for(i=1;i<=m;i++)
				{
					scanf("%d%d",&str,&end);
					map[str][end]=map[end][str]=0;   //这里必须是双向的。不能写成	map[str][end]=0。因为这里WA了两次
				}
				sum=0;
				prim();
				printf("%d\n",sum);
	}
	return 0;
}