Constructing Roads

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

题目大意：就是村庄修路，有n个村庄让你把所有村庄连接起来，并且开销最低，可以直接用最小生成树做，还有他给定了有部分路是已经修好了的。
思路：将已经修好的赋值为0，然后求出最小生成树的权值总和。
代码（c++）：

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxsize=0x3f3f3f3f;
int mp[108][108];
int book[108];
int dis[108];
int n;
int debug;
void prime(int begin)
{
	book[1]=1;
	int sum=0;
	int start=begin;
	for(int i=1;i<=n;i++)
	{
		dis[i]=mp[start][i];
	//	cout<<dis[i]<<"  ";
	}
	for(int o=1;o<=n-1;o++)
	{
		int minn=maxsize;
		for(int i=1;i<=n;i++)
		{
			if(dis[i]<minn&&book[i]==0)
			{
				start=i;
				minn=dis[i];
			}
		//	cout<<minn<<endl;
		}
			sum+=minn;
		book[start]=1;
		for(int i=1;i<=n;i++)
		{
			if(book[i]==0&&mp[start][i]<dis[i])
			dis[i]=mp[start][i];
		}
	}
	cout<<sum<<endl;
}
int main()
{
	while(cin>>n)
	{
		int debug=0;
		memset(dis,88,sizeof(dis));
		memset(book,0,sizeof(book));
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				int v;cin>>v;
				mp[i][j]=v;
			}
		}
		cin>>debug;
		for(int j=1;j<=debug;++j)
		{
			int a,b;
			cin>>a>>b;
			mp[a][b]=mp[b][a]=0;
		}
		prime(1);
	}
}