POJ 2421 Constructing Roads

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：

有n个村庄，每个点都有自己的权值， 其中已经有给定的某几条路已经建成， 问在这些路的基础上建通所有村庄的最小花费是多少。

解题思路：

裸的最小生成树， 对于给定的两点已有的通路只要置为0即可。

代码：

#include <iostream>
#include <sstream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <utility>
#include <string>
#include <cmath>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>

using namespace std;

/*tools:
 *ios::sync_with_stdio(false);
 *freopen("input.txt", "r", stdin);
 */

typedef long long ll;
typedef unsigned long long ull;
const int dir[5][2] = {0, 1, 0, -1, 1, 0, -1, 0, 0, 0};
const ll ll_inf = 0x7fffffff;
const int inf = 0x3f3f3f;
const int mod = 1000000;
const int Max = (int) 1e4;

int Map[Max][Max], vis[Max];
int n, q;

void Init() {
    for (int i = 0; i < n; ++i) {
        vis[i] = 0;
        for (int j = 0; j < n; ++j) {
            Map[i][j] = (i == j) ? 0 : inf;
        }
    }
}

void Prim() {
    int dis[Max], ans = 0;
    for (int i = 1; i <= n; ++i) {
        dis[i] = Map[1][i];
    }
    dis[1] = 0;
    for (int i = 1; i <= n; ++i) {
        int Min = inf, mark;
        for (int j = 1; j <= n; ++j) {
            if (!vis[j] && dis[j] < Min) {
                Min = dis[mark = j];
            }
        }
        vis[mark] = 1;
        ans += dis[mark];
        for (int j = 1; j <= n; ++j) {
            if (!vis[j] && dis[j] > Map[mark][j]) {
                dis[j] = Map[mark][j];
            }
        }
    }
    printf("%d\n", ans);
}

int main() {
    //freopen("input.txt", "r", stdin);

    // initialization
    scanf("%d", &n);
    Init();
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            scanf("%d", &Map[i][j]);
        }
    }
    scanf("%d", &q);
    // the road has been build
    for (int i = 0; i < q; ++i) {
        int a, b;
        scanf("%d %d", &a, &b);
        Map[a][b] = Map[b][a] = 0;
    }
    Prim();
    return 0;
}