PP and QQ

思路

删边游戏了解一下，其实就是个nim博弈吧，只是删边个数有特殊限制，

然后就是一个反nim博弈了。

删边定理：

• 遇到分叉口时，它的子树上的可操作的sg函数为所有子树节点的sg函数的异或值，

  然后这个异或值以一颗子树的形式与这个点连为一棵树，然后不断递归得到这一整棵树的sg函数

反nim博弈：

必胜满足$sg=0,nu{m}_{value==1}=even$或者$sg>0,nu{m}_{value>1}>=1$,

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}vector<int> G[110];int dfs(int rt, int fa) {int ans = 1, now = 0;for(int i : G[rt]) {if(i == fa) continue;now ^= dfs(i, rt);}return ans + now;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;while(scanf("%d", &T) != EOF) {int ans = 0, num = 0;for(int cas = 1; cas <= T; cas++) {int n; scanf("%d", &n);for(int i = 1; i <= n; i++) {G[i].clear();}for(int i = 1; i < n; i++) {int x, y;scanf("%d %d", &x, &y);G[x].pb(y);G[y].pb(x);}int temp = dfs(1, 0) - 1;ans ^= temp;if(temp > 1) num++;}if((ans && num) || (!ans && !num)) puts("PP");else puts("QQ");}return 0;
}