jzoj6826-[2020.10.17提高组模拟]隔膜【博弈论】

正题


题目大意

n∗nn*nnn的矩形,每一个人操作时如果棋盘上有一个k∗kk*kkk的矩形空地就可以选择一个点堵上。如果没有就失败了,求必胜方。


解题思路

如果场地上有一个位置堵上后即可堵上所有k∗kk*kkk的矩形那么这个点被堵住后就赢了,所以先手必胜。

如果没有这个位置,那么最后场地上剩下一个k∗kk*kkk的矩形时先手必胜,也就是如果我们找出两个不相交的k∗kk*kkk矩形那么显然先手不会去堵上这两个中的任何一个,也就是其他位置会被优先堵上。所以只要判断其他格子数量的奇偶性即可。


codecodecode

#include<cstdio>
#include<cstring>
#include<algorithm> 
using namespace std;
const int N=1100;
int n,k,a[N][N],b[N][N];
char s[N]; 
bool check(int x,int y)
{x--;y--;return (a[x+k][y+k]+a[x][y]-a[x][y+k]-a[x+k][y])==0;}
int main()
{
//	freopen("lcyrcx.in","r",stdin);
//	freopen("lcyrcx.out","w",stdout);scanf("%d%d",&n,&k);for(int i=1;i<=n;i++){scanf("%s",s+1);for(int j=1;j<=n;j++)a[i][j]=a[i-1][j]+a[i][j-1]+(s[j]=='1')-a[i-1][j-1];}int z=0;for(int i=1;i<=n-k+1;i++)for(int j=1;j<=n-k+1;j++)if(check(i,j)){b[i][j]++;b[i+k][j+k]++;b[i+k][j]--;b[i][j+k]--;z++;}bool flag=1;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++){b[i][j]+=b[i-1][j]+b[i][j-1]-b[i-1][j-1];if(b[i][j]==z)flag=0;}if(flag){if((a[n][n]-2*k*k)&1)printf("rx");else printf("yc");}else{if(z)printf("rx");else printf("yc");}
}
/*
100 20
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*/

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