2022.2.2 训练日记2 AcWing 99. 激光炸弹

题目链接：激光炸弹

题目分析：

0.二维前缀和。
1.参考蓝书22页。

code:

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 5010;

int s[N][N];//前缀和矩阵
int n, m;
int main()
{
    int cnt, R;
    scanf("%d%d", &cnt, &R);
    
    R = min(R, 5001);
    n = m = 5001;
    for(int i = 0; i < cnt; i ++)
    {
        int x, y, w;
        scanf("%d%d%d", &x, &y, &w);
        x++, y++;//把下标映射到1~5001。右移一位, 就不需要考虑边界了,
        s[x][y] += w;
    }
    //预处理前缀和
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
            s[i][j] += s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];//二维前缀和公式
    
    int res = 0;
    // 枚举所有边长是R的矩形，枚举(i, j)为右下角
    for(int i = R; i <= n; i ++)
        for(int j = R; j <= m; j ++)
            res = max(res,s[i][j] - s[i - R][j] - s[i][j - R] + s[i - R][j - R]);
    printf("%d\n", res);
    return 0;
}

总结：

1二维前缀和公式：
	1.1 s[i][j] = s[i][j] + s[i-1][j] + s[i][j - 1] - s[i-1][j-1]
	1.2 维护一个边长为R的正方形
		res = s[i][j] - s[i - r][j] - s[i][j - r] + s[i - r][j - r]

标签：cnt,前缀,int,d%,5001,99,include,2022.2,AcWing
来源： https://blog.csdn.net/qq_53244181/article/details/122771369