Constructing Roads 最小生成树

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

一道比较裸的最小生成树问题，题目的叙述容易让人理解错误题意，只要能联通就可以，并不是说只有直接相连和A-B+B-C=A-C这两种情况才能连起来，中间过程多少都可以，理解好这一点，就是一道裸题，直接上PRIM算法即可。

AC代码

#include
#include
#include
using namespace std;
int n,q,ans=0;
int Map[105][105];
int dist[105];
int vis[105];
int INF=0x3f3f3f;
int FindMinDist()
{
	int min=-1,mindist=INF;
	for(int i=1;i<=n;i++)
	{
		if(vis[i]==0&&dist[i]