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题目链接:
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> //#include <bits/stdc++.h> #include <stack> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar()); for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + ‘0‘); putchar(‘\n‘); } const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=2e6+10; const int maxn=500+10; const double eps=1e-8; int sg[N],vis[25]; inline int get_sg(int x) { mst(vis,0); //cout<<x<<endl; for(int i=19;i>=0;i--) { if(x&(1<<i)) { int temp=x; for(int j=i-1;j>=0;j--) { if(!(x&(1<<j))) { temp^=(1<<i)^(1<<j); //cout<<temp<<endl; vis[sg[temp]]=1; break; } } } } for(int i=0;i<=19;i++)if(!vis[i])return i; return 0; } inline void Init() { For(i,0,(1<<20)-1)sg[i]=get_sg(i); } int main() { Init(); int t; read(t); while(t--) { int n,a,m,ans=0; read(n); while(n--) { int sum=0; read(m); For(i,1,m) { read(a); sum|=(1<<(20-a)); } ans^=sg[sum]; } if(ans)printf("YES\n"); else printf("NO\n"); } return 0; }
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5701384.html