学步园

很简单的一道题目,就主要是吧已经修建成功的赋成map[i][j]就行了.

贴出代码:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#define inf 0x3fffffff

int N,M;// N represents the number of points,and M stands for the number of that has been built points;
int map[105][105];
int dis[105];
int visit[105];

int prim()
{
	for(int i=1;i<=N;i++)
		dis[i]=inf;
	dis[1]=0;
	for(int j=1;j<=N;j++)
	{
		int t=inf,pos;
		for(i=1;i<=N;i++)
		{
			if(!visit[i]&&t>dis[i])
			{
				t=dis[i];
				pos=i;
			}
		}
		visit[pos]=1;
		for(i=1;i<=N;i++)
		{
			if(!visit[i]&&dis[i]>map[pos][i]&&map[pos][i]!=0x3f3f3f3f)
			{
				dis[i]=map[pos][i];
			}
		}
	}	
	int temp=0;
	for(int k=1;k<=N;k++)
	{
		temp+=dis[k];
	}
	return temp;
}


int main()
{
	int a,b;
	while(scanf("%d",&N)!=EOF)
	{
		memset(map,0x3f,sizeof(map));
		memset(visit,0,sizeof(visit));
		for(int i=1;i<=N;i++)
		{
			for(int j=1;j<=N;j++)
			{
				scanf("%d",&map[i][j]);
			}
		}
		scanf("%d",&M);
		for(i=1;i<=M;i++)
		{
			scanf("%d%d",&a,&b);
			map[a][b]=0;
			map[b][a]=0;
		}
		int ans=prim();
		printf("%d\n",ans);
	}
	return 0;
}