bzoj4423 [AMPPZ2013]Bytehattan

传送门

分析

我们考虑把网格图转换为一个对偶图

然后我们每次删一条边就是将对偶图上对应的两个点用并查集连接起来

每次查询看是否联通即可

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
int n,m,fa[4000000];
inline int sf(int x){return fa[x]==x?x:fa[x]=sf(fa[x]);}
int main(){
    int i,j,k,wh=0;
    scanf("%d%d",&n,&m);
    n++;
    for(i=1;i<=n*n;i++)fa[i]=(i<=n||i>(n-1)*n||i%n==1||i%n==0)?1:i;
    for(i=1;i<=m;i++){
      int x,y,xx,yy;
      char s1[5],s2[5];
      scanf("%d%d",&x,&y);
      scanf("%s",s1);
      scanf("%d%d",&xx,&yy);
      scanf("%s",s2);
      if(wh){
        x=xx;
        y=yy;
        s1[0]=s2[0];
      }
      xx=sf(x*n+y+1);
      if(s1[0]=='N')yy=sf((x-1)*n+y+1);
        else yy=sf(x*n+y);
      if(xx!=yy){
        wh=0;
        puts("TAK");
        fa[xx]=yy;
      }else {
        wh=1;
        puts("NIE");
      }
    }
    return 0;
}