POJ-2421 Constructing Roads

Constructing Roads

Time Limit: 2000MS

Memory Limit: 65536K

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

————————————————————集训8.7的分割线————————————————————

思路：临接矩阵，Prim。Prim的过程如下：

1. 首先取出0点，加入最小生成树的点集

2. 更新数组lowcost为其它点集和0点的距离，已访问0点

3. 循环n - 1次，每次选出一个距离生成树点集最近的一个点k，加入生成树

4. 已访问k点，更新数组lowcost为其它点集和新的生成树点集的最近距离

这其中第四步，lowcost[ j ] = min(lowcost[ j ], mat[ k ][ j ])，指的是对于新加入的点k，如果有和它相连的点更接近最小生成树，则更新lowcost[ j ]为这个更近的距离。

其他要注意的，就是已经建好的路权值变为0。

P.S. 输入的编号是从1开始的，没发现……

代码如下：

/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <iostream>
#define INF 2e9
using namespace std;
/****************************************/
const int N = 105;
int mat[N][N], lowcost[N];
int n, Q;
bool vis[N];

int prim()
{   
	int sum = 0;
    for(int i = 0; i < n; i++)
        lowcost[i] = mat[0][i];//初始化为所有点与0的距离
    lowcost[0] = -1;
    int k;
    for(int i = 1; i < n; i++) {
        int mini = INF;
        for(int j = 0; j < n; j++) if(lowcost[j] != -1) {
            if(mini > lowcost[j]) {
                mini = lowcost[j];
                k = j;
            }
        }//取出权值最小的点加入生成树
        lowcost[k] = -1;
        sum += mini;
        for(int j = 0; j < n; j++) {
            lowcost[j] = min(lowcost[j], mat[k][j]);//更新两个点集之间的最小权值
        }
    }
    return sum;
}

int main()
{
    while(scanf("%d",&n)!=EOF) {
        memset(vis, 0, sizeof(vis));
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                scanf("%d", &mat[i][j]);
                mat[j][i] = mat[i][j];
            }
        }
        scanf("%d", &Q);
        while(Q--) {
            int a, b;
            scanf("%d%d", &a, &b);
            a--; b--;//这里没注意
            mat[a][b] = mat[b][a] = 0;
        }
        printf("%d\n", prim());
    }
    return 0;
}