POJ 2421--Constructing Roads【水题 && 最小生成树 && kruskal】

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Description

Input

Output

Sample Input

3
0 990 692
990 0 179
692 179 0
1
1 2

Sample Output

179

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;

int per[110];
int map[110][110];
int N, Q;
struct node{
    int u, v, w;
};
node edge[20000];

int cmp(node a, node b){
    return a.w < b.w;
}

void init(){
    for(int i = 1; i <= N; ++i)
        per[i] = i;
}

int find(int x){
    if(x == per[x])
        return x;
    return per[x] = find(per[x]);
}

bool join (int x, int y){
    int fx = find(x);
    int fy = find(y);
    if(fx != fy){
        per[fx] = fy;
        return true;
    }
    return false;
}

int main (){
    while(scanf("%d", &N) != EOF){
        int k = 0;
        for(int i = 1; i <= N; ++i)
        for(int j = 1; j <= N; ++j)
            scanf("%d", &map[i][j]);
        scanf("%d", &Q);
        while(Q--){
            int u, v;
            scanf("%d%d", &u, &v);
            map[u][v] = 0;
        }
        for(int i = 1; i <= N; ++i)
        for(int j = 1; j <= N; ++j){
            edge[k].u = i;
            edge[k].v = j;
            edge[k].w = map[i][j];
            k++;
        }
        sort(edge, edge + k, cmp);
        int sum = 0;
        init();
        for(int i = 0; i < k; ++i){
            //printf("---%d %d %d\n", edge[i].u, edge[i].v, edge[i].w);
            if(join(edge[i].u, edge[i].v))
                sum += edge[i].w;
        }
        printf("%d\n", sum);
    }
    return 0;
}

POJ 2421--Constructing Roads【水题 && 最小生成树 && kruskal】

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原文地址：http://blog.csdn.net/hpuhjh/article/details/47402855