HDU 5724-Chess-博弈论-sg函数

来源：互联网发布：淘宝培训kehuda 编辑：程序博客网时间：2024/05/18 01:59

对于每一个状态，暴力求出所有后继状态，通过SG函数求到该状态的SG值。。

这题用dfs+记忆化暴力写就好了

当时直接枚举了所有情况打表。。。存了所有后继，差点爆内存，改成short才过

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <queue>#include <map>#include <set>#include <vector>#include <iostream>#include<queue>#define LL long long#define MAX 10005using namespace std;/*int A[MAX];bool win[1<<21];void solve(){   for(int num=0;num<20;num++)   {       for(int i=0;i<20;i++)       {           int now=pow(2,i+1)-1;           sg[now]=0;       }   }}*/short idx[500+(1<<20)];short que[500+(1<<20)][25];short sg[500+(1<<20)];short cal_sg(int x){    if (sg[x]!=-1) return sg[x];    int vis[25];    memset(vis,0,sizeof vis);    for (int i=1; i<=idx[x]; i++)    {        vis[que[x][i]]=1;    }    for (int i=0;; i++)        if (vis[i]==0) return sg[x]=i;}long long tmpp=0;void BinaryRecursion(int n){ int a; a=n%2; n=n>>1; if (n==0)  ; else  BinaryRecursion(n); tmpp=tmpp*10+a;}void get(){    int vis[25];    int all=1<<20;    for (int i=0; i<all; i++)    {        int state=i;        sg[i]=cal_sg(state);        for (int j=0; j<20; j++)        {            if ((1<<j)&state)            {                int temp=state|((1<<j)-1);                temp=~temp;                if(0==temp) continue;                int low_zero=temp&-temp;  //lowbit                int father=state^(1<<j);                father=father| (low_zero);                //printf("%d ***\n",father);                if (father >=(all))                    continue;                que[father][++idx[father]]=cal_sg(state);                // que[father].push_back(state);             }        }    }}int main(){    memset(sg,-1,sizeof sg);     memset(idx,0,sizeof idx);   // freopen("out.txt","w",stdout);    get();   // sg[0]=0; int t;    cin>>t;    while(t--)    {        int n,m;        scanf("%d",&n);        int ans=0;        for (int i=1;i<=n;i++)        {            scanf("%d",&m);            int state=0;            int x;            for (int j=1;j<=m;j++)            {                scanf("%d",&x);                state=state|(1<<(20-x));            }            ans^=sg[state];        }        if (ans)        printf("YES\n");        else            printf("NO\n");    }    return 0;}

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