POJ

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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！

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