POJ - 1308

           不是并查集！

        思路：直接记录每个点被指向的次数，如果满足只有一个点被指向次数为0，并且其他点被指向次数为1，就是符合题目条件的树。

坑点：可能是空树。

AC代码

#include <cstdio>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000") 
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int> 
typedef long long LL;
const int maxn = 1000 + 5;
int vis[maxn], cnt[maxn];

int main() {
	int x, y, kase = 1;
	while(scanf("%d%d", &x, &y) == 2) {
		if(x == -1 && y == -1) break;
		if(x == 0 && y == 0) { //空树 
			printf("Case %d is a tree.\n", kase++);
			continue; 
		}
		memset(vis, 0, sizeof(vis));
		memset(cnt, 0, sizeof(cnt));
		do{
			vis[x] = vis[y] = 1;
			cnt[y]++;
			scanf("%d%d", &x, &y);
		}while(x != 0 || y != 0);
		int flag = 1, tot = 0;
		for(int i = 1; i < maxn; ++i) {
			if(vis[i]) {
				if(!cnt[i]) ++tot;
				if(cnt[i] > 1 || tot > 1) {
					flag = 0;
					break;
				}
			}
		}
		if(flag) printf("Case %d is a tree.\n", kase++);
		else printf("Case %d is not a tree.\n", kase++);
	}
	return 0;
}

如有不当之处欢迎指出！