【洛谷P4514】上帝造题的七分钟

Description

给定一个矩阵，要求实现区间修改，区间求和的操作

Solution

二维树状数组的模板题，类比一维，我们依旧利用差分的思想完成。

首先，运用简单的容斥思想，二维前缀和sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+a[i][j]

查询左上角点(x1,y1)，右下角点(x2,y2)的区间和则是sum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]+sum[x1-1][y1-1]

那么我们用树状数组维护差分数组即可。

具体地，根据差分数组的含义，对于点(x,y)，他的前缀和是

$$sumlimits_{i=1}^{x}sumlimits_{j=1}^{y}sumlimits_{k=1}^{i}sumlimits_{l=1}^{j}{d[k][l]}$$

化简，得

$$sumlimits_{i=1}^{x}sumlimits_{j=1}^{y}{d[i][j](x-i+1)(y-j+1)}$$

分解，得

$$sumlimits_{i=1}^{x}sumlimits_{j=1}^{y}{d[i][j](xy+x+y+1)-d[i][j]i(y+1)-d[i][j]j(x+1)+d[i][j]ij}$$

根据这个式子，我们维护四个树状数组d[i][j],d[i][j]i,d[i][j]j,d[i][j]ij即可

Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline int read() {
    int ret = 0, op = 1;
    char c = getchar();
    while (!isdigit(c)) {
        if (c == '-') op = -1; 
        c = getchar();
    }
    while (isdigit(c)) {
        ret = ret * 10 + c - '0';
        c = getchar();
    }
    return ret * op;
}
int n, m;
struct BIT {
    int a[2050][2050];
    inline int lowbit(int x) {
        return x & (-x);
    }
    int query(int x, int y) {
        int ret = 0;
        for (register int i = x; i >= 1; i -= lowbit(i))
            for (register int j = y; j >= 1; j -= lowbit(j))
                ret += a[i][j];
        return ret;
    }
    void add(int x, int y, int val) {
        for (register int i = x; i <= n; i += lowbit(i))
            for (register int j = y; j <= m; j += lowbit(j))
                a[i][j] += val;
    }
} A, Ai, Aj, Aij;
void add(int x, int y, int val) {
    A.add(x, y, val);
    Ai.add(x, y, val * x);
    Aj.add(x, y, val * y);
    Aij.add(x, y, val * x * y);
}
int query(int x, int y) {
    return A.query(x, y) * (x * y + x + y + 1) - Ai.query(x, y) * (y + 1) - Aj.query(x, y) * (x + 1) + Aij.query(x, y);
}
int main() {
    getchar();
    n = read(); m = read();
    char op[3];
    while (~scanf("%s", op)) {
        int x1 = read(), y1 = read(), x2 = read(), y2 = read();
        if (op[0] == 'L') {
            int z = read();
            add(x1, y1, z);
            add(x1, y2 + 1, -z);
            add(x2 + 1, y1, -z);
            add(x2 + 1, y2 + 1, z);
        }
        else printf("%d
", query(x2, y2) - query(x1 - 1, y2) - query(x2, y1 - 1) + query(x1 - 1, y1 - 1));
    }
    return 0;
}