CF869E The Untended Antiquity

题目描述

$\text{Adieu l'ami}$.

Koyomi is helping Oshino, an acquaintance of his, to take care of an open space around the abandoned Eikou Cram School building, Oshino's makeshift residence.

The space is represented by a rectangular grid of $n \times m$ cells, arranged into $n$ rows and $m$ columns. The $c$ -th cell in the $r$ -th row is denoted by $(r,c)$ .

Oshino places and removes barriers around rectangular areas of cells.

Specifically, an action denoted by " $1\ r_{1}\ c_{1}\ r_{2}\ c_{2}$" means Oshino's placing barriers around a rectangle with two corners being $(r_{1},c_{1})$ and $(r_{2},c_{2})$ and sides parallel to squares sides.

Similarly, " $2\ r_{1}\ c_{1}\ r_{2}\ c_{2}$ " means Oshino's removing barriers around the rectangle. Oshino ensures that no barriers staying on the ground share any common points, nor do they intersect with boundaries of the $n \times m$ area.

Sometimes Koyomi tries to walk from one cell to another carefully without striding over barriers, in order to avoid damaging various items on the ground. " $3\ r_{1}\ c_{1}\ r_{2}\ c_{2}$" means that Koyomi tries to walk from $(r_{1},c_{1})$ to $(r_{2},c_{2})$ without crossing barriers.

And you're here to tell Koyomi the feasibility of each of his attempts.

输入输出格式

输入格式：

The first line of input contains three space-separated integers $n$ , $m$ and $q$ ( $1<=n,m<=2500$ , $1<=q<=100000$ ) — the number of rows and columns in the grid, and the total number of Oshino and Koyomi's actions, respectively.

The following q q lines each describes an action, containing five space-separated integers $t$ , $r_{1}$ , $c_{1}$ , $r_{2}$ , $c_{2}$ ( $1<=t<=3$ , $1<=r_{1} , r_{2}<=n$ , $1<=c_{1},c_{2}<=m$ ) — the type and two coordinates of an action. Additionally, the following holds depending on the value of t t :

If $t=1$ : $2<=r_{1}<=r_{2}<=n-1$ , $2<=c_{1}<=c_{2}<=m-1$ ;

If $t=2$ : $2<=r_{1}<=r_{2}<=n-1$ , $2<=c_{1}<=c_{2}<=m-1$ , the specified group of barriers exist on the ground before the removal.

If $t=3$ : no extra restrictions.

输出格式：

For each of Koyomi's attempts (actions with $t=3$ ), output one line — containing "Yes" (without quotes) if it's feasible, and "No" (without quotes) otherwise.

这题有几个要注意的条件：围墙不重叠不交叉不堵边界

可以对一个围墙里面的矩形进行染色。

查询看颜色是否一样。

具体来说，拿二维树状数组维护单点问区间改，差分一下

拿了随机数长整型自动取模wa，模大质数wa，听说不能随机。。

然后试了试，维护异或和就过了

注意维护异或和差分的写法。

Code:

#include <cstdio>

#include <cstdlib>

#include <ctime>

#define ll long long

const int N=2505;

int n,m,q;

ll s[N][N],tag[N][N];

void add(int x,int y,ll d)

{

    for(int i=x;i<=n;i+=i&-i)

        for(int j=y;j<=m;j+=j&-j)

            s[i][j]^=d;

}

ll query(int x,int y)

{

    ll sum=0;

    for(int i=x;i;i-=i&-i)

        for(int j=y;j;j-=j&-j)

            sum^=s[i][j];

    return sum;

}

int main()

{

    srand(19260817);

    scanf("%d%d%d",&n,&m,&q);

    for(int r1,c1,r2,c2,op,i=1;i<=q;i++)

    {

        scanf("%d%d%d%d%d",&op,&r1,&c1,&r2,&c2);

        if(op==1)

        {

            tag[r1][c1]=rand()*rand();

            add(r1,c1,tag[r1][c1]);

            add(r2+1,c2+1,tag[r1][c1]);

            add(r1,c2+1,tag[r1][c1]);

            add(r2+1,c1,tag[r1][c1]);

        }

        else if(op==2)

        {

            add(r1,c1,tag[r1][c1]);

            add(r2+1,c2+1,tag[r1][c1]);

            add(r1,c2+1,tag[r1][c1]);

            add(r2+1,c1,tag[r1][c1]);

        }

        else

        {

            if(query(r1,c1)==query(r2,c2)) puts("Yes");

            else puts("No");

        }

    }

    return 0;

}

2018.10.11