ＰＯＪUbiquitous Religions（非常棒的并查集入门题目）

#include <iostream>#include <cstdio>using namespace std;int f[50010];int sum = 0;//Accepted360K313MSint find1(int x) {    if(f[x] != x) {        f[x] = find1(f[x]);    }    return f[x];}void merge1(int a, int b) {    int f1 = find1(a);    int f2 = find1(b);    if(f1 != f2) {        f[f2] = f1;        sum--;    }}int main(){    int n, m, p = 1;    while(scanf("%d%d", &n, &m) != EOF) {        if( n == 0 && m == 0) { break; }        for(int i = 1; i <= n; i++) {            f[i] = i;        }        sum = n;        for(int i = 1; i <= m; i++) {            int a, b;            scanf("%d%d", &a, &b);            merge1(a, b);        }        printf("Case %d: %d\n", p++, sum);    }    return 0;}／／另一版本，大概思路相似，仅仅就并的时候，少的像多的并。#include <iostream>#include <cstdio>#include <cstring>using namespace std;//Accepted556K297MSstruct student {    int index;    int num;    student(){        num = 1;    }};student f[50010];///刚刚开始数组开小了，runtime error!int n, m, sum = 0;int find1(int x) {//查找    if(f[x].index != x) {       f[x].index = find1(f[x].index);    }    return f[x].index;}void merge1(int a, int b){ //合并    int f1 = find1(f[a].index);    int f2 = find1(f[b].index);    if(f1 != f2) {        if(f[f1].num >= f[f2].num) {            f[f1].num += f[f2].num;            f[f2].index = f1;        }        else {            f[f2].num += f[f1].num;            f[f1].index = f2;        }        sum--;    }}void init() {//初始化    sum = n;    for(int i = 1; i <= n; i++) {        f[i].index = i;    }}int main(){    int p = 1;    while(scanf("%d%d", &n, &m) != EOF) {        if(!n && !m) { break;}        init();        for(int i = 1; i <= m; i++) {            int a, b;            scanf("%d%d", &a, &b);            merge1(a, b);        }        printf("Case %d: %d\n", p++, sum);    }    return 0;}