HDU 3590 PP and QQ(反nim博弈，删边游戏)

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 s g = 0 , n u m v a l u e = = 1 = e v e n sg = 0, num_{value == 1} = even sg=0,numvalue==1​=even或者 s g > 0 , n u m v a l u e > 1 > = 1 sg > 0, num_{value > 1} >= 1 sg>0,numvalue>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 | 1

using 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;
}

本文地址：https://blog.csdn.net/weixin_45483201/article/details/108859165