CF835C Star sky (二維前綴和）

CF835C Star sky
洛谷鏈接

題目描述
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.
輸入格式
The first line contains three integers n , q , c ( 1<=n,q<=10 ^5, 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<=10^9
, 1<=x1i,x2i<=100 , 1<=y1i,y2i<=100 ) — the moment of the i -th view and the coordinates of the viewed rectangle.

輸出格式
For each view print the total brightness of the viewed stars.

輸入輸出樣例
輸入 #1
2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5
輸出 #1
3
0
3

輸入 #2
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
輸出 #2
3
3
5
0
說明/提示
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 .

Solution

二維前綴和 + 加一維表示亮度

不用更新，所以不需要使用樹狀數組維護，直接做二維前綴和即可

star[i][j][k] 表示（1,1） 到（i，j）中亮度爲k的星星的數量。

代碼

#include<cmath>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
const int SZ = 100 + 5;

int star[SZ][SZ][12];
int n,q,c;

int main()
{
    int x,y,s,t,xx,yy;
    scanf("%d%d%d",&n,&q,&c);
    for(int i = 1;i <= n;i ++ )
    {
        scanf("%d%d%d",&x,&y,&s);
        star[x][y][s] ++ ;
    }
    for(int i = 1;i <= 100;i ++ )
        for(int j = 1;j <= 100;j ++ )
            for(int k = 0;k <= c;k ++ )
            {
                 star[i][j][k] += star[i - 1][j][k] + star[i][j - 1][k] - star[i - 1][j - 1][k];
            }
    
    for(int i = 1;i <= q;i ++ )
    {
        int ans = 0;
        scanf("%d%d%d%d%d",&t,&x,&y,&xx,&yy);
        for(int k = 0;k <= c;k ++ )
        {
            ans += (k + t) % (c + 1) * (star[xx][yy][k] - star[xx][y - 1][k] - star[x - 1][yy][k] + star[x - 1][y - 1][k]);
        }
        printf("%d\n",ans);
    }
    //system("pause");
    return 0;
}

2020.4.2