POJ Ubiquitous Religions（非常棒的并查集入门题目）
[cpp]#include <iostream>
#include <cstdio>
using namespace std;
int f[50010];
int sum = 0;
//Accepted 360K 313MS
int 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;
//Accepted 556K 297MS
struct 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;
}
补充：软件开发 , C++ ,
