BZOJ4423 AMPPZ2013Bytehattan（并查集）

时间：2021-07-07 22:24:37

　　判断网格图中某两点是否被割开，可以将割边视为边区域视为点，转化为可切割这两点的区域是否连通。于是每次判断使两个区域连通后是否会形成环（边界视为连通），若是则说明被两点被割开。并查集维护。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

using namespace std;

int read()

{

    int x=,f=;char c=getchar();

    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}

    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();

    return x*f;

}

char getc(){char c=getchar();while (c!='N'&&c!='E') c=getchar();return c;}

#define N 1510

int n,m,fa[N*N],last=;

int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}

int trans(int x,int y)

{

    if (x==||x==n||y==||y==n) return ;

    return (x-)*(n-)+y;

}

int main()

{

#ifndef ONLINE_JUDGE

    freopen("bzoj4423.in","r",stdin);

    freopen("bzoj4423.out","w",stdout);

    const char LL[]="%I64d\n";

#else

    const char LL[]="%lld\n";

#endif

    n=read(),m=read();

    for (int i=;i<=n*n;i++) fa[i]=i;

    while (m--)

    {

        int x=read(),y=read();char c=getc();

        if (last==) read(),read(),getc();

        else x=read(),y=read(),c=getc();

        int p=x-(c=='N'),q=y-(c=='E');

        int u=find(trans(p,q)),v=find(trans(x,y));

        last=find(u)==find(v);

        fa[u]=v;

        if (last==) printf("TAK\n");

        else printf("NIE\n");

    }

    return ;

}

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