POJ

题目：有一棵无向树，从某个结点出发，两个人轮流移动，走过的结点不能再走，不能移动的人输。判断先手必胜还是先手必败，如果先手必胜，输出必胜到达的节点。

思路：利用NP状态定理，把起点的SG值算出来。

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=1e3+5;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;        if(get_sg(v,u)==0)            sg[u]=1;    }    return sg[u];}int main(){    int n,start,u,v;    while(~scanf("%d%d",&n,&start)){        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);        }        int ans=get_sg(start,0);        if(ans>0){            int id=n+1;            for(int i=0;i<G[start].size();i++){                int v=G[start][i];                if(sg[v]==0)                    id=min(id,v);            }            printf("First player wins flying to airport %d\n",id);        }        else{            printf("First player loses\n");        }    }    return 0;}