Codeforces Round #439 (Div. 2) E. The Untended Antiquity 二维树状数组 随机化

E. The Untended Antiquity

time limit per test
2 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

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 × 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 r1 c1 r2 c2"
means Oshino's placing barriers around a rectangle with two corners being (r1, c1) and (r2, c2) and
sides parallel to squares sides. Similarly, "2 r1 c1 r2 c2"
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 × 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 r1 c1 r2 c2"
means that Koyomi tries to walk from (r1, c1) to (r2, c2) without
crossing barriers.

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

Input

The first line of input contains three space-separated integers n, m and q (1 ≤ n, m ≤ 2 500, 1 ≤ q ≤ 100 000)
— the number of rows and columns in the grid, and the total number of Oshino and Koyomi's actions, respectively.

The following q lines each describes an action, containing five space-separated integers t, r1, c1, r2, c2 (1 ≤ t ≤ 3, 1 ≤ r1, r2 ≤ n, 1 ≤ c1, c2 ≤ m)
— the type and two coordinates of an action. Additionally, the following holds depending on the value of t:

If t = 1: 2 ≤ r1 ≤ r2 ≤ n - 1, 2 ≤ c1 ≤ c2 ≤ m - 1;

If t = 2: 2 ≤ r1 ≤ r2 ≤ n - 1, 2 ≤ c1 ≤ c2 ≤ m - 1,
the specified group of barriers exist on the ground before the removal.

If t = 3: no extra restrictions.

Output

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.

Examples

input

5 6 5
1 2 2 4 5
1 3 3 3 3
3 4 4 1 1
2 2 2 4 5
3 1 1 4 4

output

No
Yes

input

2500 2500 8
1 549 1279 1263 2189
1 303 795 1888 2432
1 2227 622 2418 1161
3 771 2492 1335 1433
1 2017 2100 2408 2160
3 48 60 798 729
1 347 708 1868 792
3 1940 2080 377 1546

output

No
YesNo

Note

For the first example, the situations of Koyomi's actions are illustrated below.



考试时问了nerther_nor题解，然后树状数组写错了。。

最后贴了份代码，然后光荣skip，要不就基本上紫了.mmp

好了说题解：

每次操作把矩形异或一个随机值，然后每次查询两个点的权值一不一样就行了

具体还是很简单的

就是矩形(x,y),(a,b) 在(x,y)(a+1,b+1),(x,b+1),(a+1,y)打标记

查询二维前缀异或和就行了

然后写树状数组手懒，query粘的modify，然后减号还是加号。泪

#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<string>
#include<bitset>
#include<queue>
#include<set>
#include<map>
using namespace std;

typedef long long ll;

inline ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return x*f;
}
void print(ll x)
{if(x<0)putchar('-'),x=-x;if(x>=10)print(x/10);putchar(x%10+'0');}

const int N=2510;

int n,m;

ll bit

,v

;

inline void modify(int x,int y,ll val)
{for(;x<=n;x+=(x&-x))for(int i=y;i<=m;i+=(i&-i))bit[x][i]^=val;}

inline ll query(int x,int y)
{ll res=0;for(;x;x-=(x&-x))for(int i=y;i;i-=(i&-i))res^=bit[x][i];return res;}

int main()
{
n=read(),m=read();int Q=read();
register int x,y,a,b,opt;
while(Q--)
{
opt=read();
x=read();y=read();a=read();b=read();
if(opt==1)
{
if(x>a)swap(a,x);if(y>b)swap(b,y);
ll now=ll(rand())<<48ll|ll(rand())<<34ll|(rand())<<15ll|rand();
v[x][y]=now;
modify(x,y,now);modify(x,b+1,now);modify(a+1,b+1,now);modify(a+1,y,now);
}
else if(opt==2)
{
if(x>a)swap(a,x);if(y>b)swap(b,y);
ll now=v[x][y];
modify(x,y,now);modify(x,b+1,now);modify(a+1,b+1,now);modify(a+1,y,now);
}
else {query(x,y)^query(a,b)?puts("No"):puts("Yes");}
}
return 0;
}
/*
5 6 5 1 2 2 4 5 1 3 3 3 3 3 4 4 1 1 2 2 2 4 5 3 1 1 4 4
*/