标签：代码示例 rip upper cto cpp ide 代码 name wan
一维前缀和 ：
这个优化 ， 可以在 O （1） 的时间内计算出一个序列的和 ，
二维前缀和 ：
对于一个矩阵 ， 也可以在 O （1） 的时间内计算出矩阵 （x1~x2）( y1 ~ y2 ) 的和 。
sum[ i ] [ j ] 表示矩阵 1 ~ i ， 1 ~ j 的和 ， 那么由容斥原理知 sum[ 0 ] [ j ] 和 sum [ i ] [ 0 ] 均为 0 。
则 s[ x1 ~ x2 ] [ y1 ~ y2 ] = sum[ x2 , y2 ] + sum [ x1 - 1 ] [ y1 - 1 ] - sum [ x1 - 1 ][ y2 ] - sum [ x2 ] [ y1-1 ] 。
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.
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.
For each view print the total brightness of the viewed stars.
2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5
3
0
3
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
3
3
5
0
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.
如果这题是遍历 所有的点 ， 那么一定会超时 ， 由于灯亮度变化范围较小 ， 可以借助二维前缀和 ， 将一定范围内不同亮度的灯全部合并到一起 ， 最后对于 每次询问 ，只需要遍历所有的亮度即可 。
代码示例 ：
/*
* Author: renyi
* Created Time: 2017/8/29 9:23:49
* File Name:
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <time.h>
using namespace std;
const int maxint = -1u>>1;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define ll long long
int dp[105][105][15] ;
int main() {
int n , q , c ;
int x , y , c1 ;
while ( ~scanf ( "%d%d%d" , &n , &q , &c) ) {
memset ( dp , 0 , sizeof(dp) ) ;
for ( int i = 0 ; i < n ; i++ ) {
scanf ( "%d%d%d" , &x , &y , &c1 ) ;
dp[x][y][c1]++ ;
}
for ( int i = 1 ; i <= 100 ; i++ ) {
for ( int j = 1 ; j <= 100 ; j++ ) {
for ( int k = 0 ; k <= 10 ; k++ ) {
dp[i][j][k] += dp[i-1][j][k]+dp[i][j-1][k]-dp[i-1][j-1][k] ;
}
}
}
int t , x1 , y1 , x2 , y2 ;
while ( q-- ) {
scanf ( "%d%d%d%d%d" , &t , &x1 , &y1 , &x2 , &y2 ) ;
int sum = 0 ;
for ( int i = 0 ; i <= c ; i++ ) {
int s = dp[x2][y2][i] + dp[x1-1][y1-1][i] - dp[x1-1][y2][i] - dp[x2][y1-1][i] ;
int l = ( t + i ) % ( c + 1 ) ;
sum += s * l ;
}
printf ( "%d\n" , sum ) ;
}
}
return 0;
}
标签：代码示例 rip upper cto cpp ide 代码 name wan
原文地址：http://www.cnblogs.com/ccut-ry/p/7446958.html
