HDU

题目：有一棵树，每一次操作有两步，第一步选择一条边删除，第二步把没有和根相连的边和点全部删除。不能进行操作的人输。

思路：每一个节点的sg值都是其子节点的sg值加1的异或和

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f//0x3f3f3f3fconst int maxn=1e5+50;vector<int> G[maxn];int sg[maxn];int get_sg(int u,int fa){    if(sg[u]!=-1)        return sg[u];    sg[u]=0;    for(int i=0;i<G[u].size();i++){        int v=G[u][i];        if(v==fa) continue;        get_sg(v,u);        sg[u]=(sg[u]^(1+sg[v]));    }    return sg[u];}int main(){    int T,n,u,v;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        for(int i=1;i<=n;i++){            G[i].clear();            sg[i]=-1;        }        for(int i=1;i<n;i++){            scanf("%d%d",&u,&v);            G[u].push_back(v);            G[v].push_back(u);        }        get_sg(1,0);        if(sg[1]>0) printf("Alice\n");        else printf("Bob\n");    }    return 0;}