poj1251 Jungle Roads

标签：最小生成树

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题目链接

/*    题意：输出最小生成树的权值和。每组数据给出节点个数n，以下n-1行，          开头为起始节点，及与它相连的节点数目m，随后m组数据表示和它相连的节点以及边权值(用c++处理空格和换行符)。    思路：Kruskal。*/#include <stdio.h>#include <iostream>#include <algorithm>using namespace std;#define maxn 30typedef struct{    int x, y;    int w;}edge;edge e[maxn * maxn / 2];int rank_[maxn], father[maxn], sum;int cmp(edge a, edge b){  //    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, m, k, weight, i, j;    char a, b;    while(scanf("%d", &n) && n){        k = 0, sum = 0;        for(i = 0; i < n - 1; i++){            cin >> a >> m, make_set(a - 'A');            for(j = 0; j < m; j++){                cin >> b >> weight, make_set(b - 'A');                e[k].x = a - 'A', e[k].y = b - 'A';                e[k++].w = weight;            }        }        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;}