Codeforces 835C - Star sky 【预处理二维前缀和】

最新推荐文章于 2018-07-28 20:14:21 发布

Wwayne_

最新推荐文章于 2018-07-28 20:14:21 发布

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分类专栏： ACM codeforces 前缀和

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The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.

You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.

A star lies in a rectangle if it lies on its border or lies strictly inside it.

Input

The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.

The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si (1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.

The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.

Output

For each view print the total brightness of the viewed stars.

   Examples 
 

     input 
    
      Copy 
    
2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5

     output 
    
      Copy 
    
3
0
3

     input 
    
      Copy 
    
3 4 5
1 1 2
2 3 0
3 3 1
0 1 1 100 100
1 2 2 4 4
2 2 1 4 7
1 50 50 51 51

     output 
    
      Copy 
    
3
3
5
0

Note

Let's consider the first example.

At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.

At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.

At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.

【题意】

有一片100*100的星空，上面有n颗星星，每个星星有一个亮度，且在0~C范围内周期性变化，现在给出q个查询，每个查询给出时间和一个矩形，求在该时间时矩形内星星的亮度和。

【思路】：

学树状数组的时候遇到的，一开始以为是个二维树状数组题，结果发现不需要更新数组，直接预处理前缀和就好了。

我们用sum[i][j][k]来表示左下角（0，0），右上角为（i，j），亮度为k的矩形星星总数，迭代维护sum[i][j][k]的数量，有点Dp的意思，状态转移方程

sum[i][j][k] += sum[i-1][j][k] + sum[i][j-1][k] - sum[i-1][j-1][k]，方程中用到的容斥的思想，画图就很好理解了。

对于q个询问，求出每个矩形星星的亮度总和，再用系数（t+i）%(c+1)将时间转化到t时刻，累和得出结果

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
const int Maxn = 1e5 + 6;
int n,q,c;
int sum[109][109][19];
using namespace std;
void init(){
	for (int i = 1; i <= 100; i++)
		for (int j = 1; j <=  100; j++)
			for (int k = 0; k <= 10; k++){
				sum[i][j][k] += sum[i-1][j][k] + sum[i][j-1][k] - sum[i-1][j-1][k];
			}
}
int main ()
{
	scanf("%d%d%d",&n,&q,&c);
	for (int i = 1; i <= n; i++) {
		int x,y,t;
		scanf("%d%d%d",&x,&y,&t);
		sum[x][y][t]++;
	}
	init();
	for (int i = 1; i <= q; i++){
		int cnt = 0;
		int t,x1,y1,x2,y2;
		scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);
		for (int i = 0; i <= c; i++){
			int k = (t+i)%(c+1);
			int p = sum[x2][y2][i]+sum[x1-1][y1-1][i]-sum[x1-1][y2][i]-sum[x2][y1-1][i];
			cnt += k*p;
		}
		printf("%d\n",cnt);
	}
}