CF869 E 二维BIT

Posted 2020-10-10 ( m Lweleth)

1代表建一个屏障，2代表去掉一个屏障，3询问是否两点相通。

仿造一维询问是否在同一区间的问题扩展到二维，树状数组维护区间标记即可，标记值可以直接2500进制不会爆LL。

 

/** @Date    : 2017-10-09 21:03:38
  * @FileName: 869E 二维树状数组.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([email protected])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

LL C[2510][2510];
LL n, m, q;
void add(LL x, LL y, LL val)
{
	while(x <= n)
	{
		LL ty = y;//别直接用y阿...
		while(ty <= m)
		{
			C[x][ty] += val;
			ty += ty & (-ty);
		}
		x += x & (-x);
	}
}

void fill(LL x1, LL y1, LL x2, LL y2, LL v)
{
	add(x1, y1, v);
	add(x2 + 1, y2 + 1, v);
	add(x1, y2 + 1, -v);
	add(x2 + 1, y1, -v);
}

LL query(LL x, LL y)
{
	LL ans = 0;
	while(x > 0)
	{
		LL ty = y;
		while(ty > 0)
		{
			ans += C[x][ty];
			ty -= ty & (-ty);
		}
		x -= x & (-x);
	}
	return ans;
}

int main()
{
	MMF(C);
	cin >> n >> m >> q;
	LL t, x1, x2, y1, y2;
	for(int i = 0; i < q; i++)
	{
		scanf("%lld%lld%lld%lld%lld", &t, &x1, &y1, &x2, &y2);
		LL val = x1 + y1*2500LL + x2*2500LL*2500LL + y2*2500LL*2500LL*2500LL;
		//cout << val << endl;
		if(t <= 2)
			fill(x1, y1, x2, y2, val*(t==1?1:-1));
		else
			printf("%s\n", (query(x1, y1)==query(x2, y2)?"Yes":"No") );
	}
    return 0;
}