最小生成树----特殊已经有路产生要我们链接未有路的

Constructing Roads
描述：
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.
输入：
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.
输出：
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.
样例输入：
3
0 990 692
990 0 179
692 179 0
1
1 2
复制
样例输出：
179

解决方法很简单：因为我们不想让已经有路的节点再连上，我们可以直接让已经有的两个节点之间road[a][b]=0:

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 110, INF = 0x3f3f3f3f;
int g[N][N], road[N][N], n, dist[N];
bool st[N], havebu[N];
int prim()
{
    int ans = 0;
    for (int i = 0; i < n; i++) //遍历每一个点，使其得到优化
    {
        int t = -1;
        for (int j = 1; j <= n; j++) //找到最小的边所属的那个点:
        {
            if (!st[j] && (t == -1 || dist[t] > dist[j]))
            {
                t = j;
            }
        }
        if (i)
            ans += dist[t];
        st[t] = true;
        for (int j = 1; j <= n; j++)
        {
            dist[j] = min(dist[j], road[t][j]);
        }
    }
    return ans;
}
int main()
{
    scanf("%d", &n);
    int len;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            scanf("%d", &road[i][j]);
        }
    }
    int q;
    memset(g, 0x3f, sizeof(g));
    memset(st, false, sizeof(st));
    memset(dist, 0x3f, sizeof(dist));
    scanf("%d", &q);
    int a, b;
    for (int i = 1; i <= q; i++)
    {
        scanf("%d%d", &a, &b);
        road[a][b]=road[b][a]=0;//解决之处
    }
    int ans = prim();
    printf("%d", ans);
    return 0;
}