HDU 1102 Constructing Roads（prim最小生成树）

Constructing Roads

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26072    Accepted Submission(s): 9998

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

看完这个题没想起来怎么做，再一看文档分类基础最小生成树，基础。。。。心里凉了许多，心想先敲着prim模板吧，结果没敲二行，想起来怎么做了。将已建成的边权值赋为0就行了。

#include
#include 
using namespace std;
int mat[105][105];
#define inf 0x3f3f3f3f
int dist[105];
int vis[105];
int n,v0;
void prim()//prim模板 
{
	int sum=0;
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
	{
		dist[i]=mat[v0][i];
	}
	dist[v0]=0;
	vis[v0]=1;
	for(int i=1;imat[pos][j])
			dist[j]=mat[pos][j];
		}
	}
	printf("%d\n",sum);
	return;
}
int main()
{
	int m;
	while(~scanf("%d",&n))
	{
		for(int i=1;i<=n;i++)//各端点连接的权值 
		{
			for(int j=1;j<=n;j++)
			{
				scanf("%d",&mat[i][j]);
				if(i==j)//题中相同端点设为0，这里要赋值inf 
				{
					mat[i][j]=inf;
				}
			}
		}
		scanf("%d",&m);
		int a,b;
		for(int i=0;i