poj1258 Agri-Net

poj1258 Agri-Net

标签：最小生成树

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题目链接

/*
    题意：给出n*n的无向图的邻接矩阵，输出最小生成树的权值和。
    思路：Kruskal。
    优化：无向图矩阵是对称的，只保存上三角。到最后k的值就是边数，由循环n*n-->k次。
*/
#include <stdio.h>
#include <algorithm>
using namespace std;

#define maxn 105

typedef struct{
    int x, y;
    int w;
}edge;

edge e[maxn * maxn / 2];
int rank_[maxn], father[maxn], sum;

int cmp(edge a, edge b){  //按权值非降序排序
    return a.w < b.w;
}

void make_set(int x){
    father[x] = x;
    rank_[x] = 0;
}

int find_set(int x){
    return  x != father[x] ? find_set(father[x]) : father[x];
}

void union_set(int x, int y, int w){
    if(x == y)  return ;
    if(rank_[x] > rank_[y])  father[y] = x;
    else{
        if(rank_[x] == rank_[y])  rank_[y]++;
        father[x] = y;
    }
    sum += w;  //
}

int main(){
    int n, k, i, j, no;

    while(scanf("%d", &n) != EOF){
        k = 0, sum = 0;
        for(i = 0; i < n; i++){
            make_set(i);
            for(j = 0; j < n; j++)
                if(j < i){  //只读取下三角
                    e[k].x = i, e[k].y = j;
                    scanf("%d", &e[k++].w);
                }
                else  scanf("%d", &no);
        }
        sort(e, e + k, cmp);
        for(i = 0; i < k; i++)  union_set(find_set(e[i].x), find_set(e[i].y), e[i].w);
        printf("%d\n", sum);
    }

    return 0;
}