克鲁斯卡尔Kruskal

寻找直接或间接连通所有城市的,最小费用的道路集合的问题,就是寻找最小生成树的问题.

#include <stdio.h>#include <stdlib.h>#include <malloc.h>#define MAX 100/* 定义边(x,y)，权为w */typedef struct{    int x, y;    int w;}edge;edge e[MAX];/* rank[x]表示x的秩 */int rank[MAX];/* father[x]表示x的父节点 */int father[MAX];int sum; //typedef int (*cmptr)(const int a, const int b);int cmp(const void*a,const void*b){    return (*(edge*)a).w>(*(edge*)b).w?1:-1;}/*堆排序*//* 初始化集合 */void Make_Set(int x){    father[x] = x;    rank[x] = 0;}/* 查找x元素所在的集合,回溯时压缩路径 */int Find_Set(int x){    if (x != father[x])    {        father[x] = Find_Set(father[x]);    }    return father[x];}/* 合并x,y所在的集合 */void Union(int x, int y, int w){    if (x == y) return;    /* 将秩较小的树连接到秩较大的树后 */    if (rank[x] > rank[y])    {        father[y] = x;    }    else    {        if (rank[x] == rank[y])        {            rank[y]++;        }        father[x] = y;    }    sum += w;}/* 主函数 */int main(){int i, n;    int x, y;    char chx, chy;    for(i=1;i<=26;i++)    Make_Set(i);    scanf("%d", &n);    getchar();    for (i = 1; i <=n; i++)    {        scanf("%c%c", &chx, &chy);        getchar();        scanf("%d",&e[i].w);        getchar();        e[i].x = chx - 'a'+1;        e[i].y = chy - 'a'+1;    }    qsort(e,n,sizeof(e[0]),cmp);    sum = 0;    for (i = 1; i <=n; i++)    {        x = Find_Set(e[i].x);        y = Find_Set(e[i].y);        if (x != y)        {            printf("%c - %c : %d\n",e[i].x + 'a'-1, e[i].y + 'a'-1, e[i].w);            Union(x, y, e[i].w);        }    }    printf("Total:%d\n", sum);    return 0;   }