【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

$$\sum\limits_{i=1}^{x}\sum\limits_{j=1}^{y}\sum\limits_{k=1}^{i}\sum\limits_{l=1}^{j}{d[k][l]}$$

化简，得

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

分解，得

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

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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 typedef long long ll;
 7 inline int read() {
 8     int ret = 0, op = 1;
 9     char c = getchar();
10     while (!isdigit(c)) {
11         if (c == '-') op = -1; 
12         c = getchar();
13     }
14     while (isdigit(c)) {
15         ret = ret * 10 + c - '0';
16         c = getchar();
17     }
18     return ret * op;
19 }
20 int n, m;
21 struct BIT {
22     int a[2050][2050];
23     inline int lowbit(int x) {
24         return x & (-x);
25     }
26     int query(int x, int y) {
27         int ret = 0;
28         for (register int i = x; i >= 1; i -= lowbit(i))
29             for (register int j = y; j >= 1; j -= lowbit(j))
30                 ret += a[i][j];
31         return ret;
32     }
33     void add(int x, int y, int val) {
34         for (register int i = x; i <= n; i += lowbit(i))
35             for (register int j = y; j <= m; j += lowbit(j))
36                 a[i][j] += val;
37     }
38 } A, Ai, Aj, Aij;
39 void add(int x, int y, int val) {
40     A.add(x, y, val);
41     Ai.add(x, y, val * x);
42     Aj.add(x, y, val * y);
43     Aij.add(x, y, val * x * y);
44 }
45 int query(int x, int y) {
46     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);
47 }
48 int main() {
49     getchar();
50     n = read(); m = read();
51     char op[3];
52     while (~scanf("%s", op)) {
53         int x1 = read(), y1 = read(), x2 = read(), y2 = read();
54         if (op[0] == 'L') {
55             int z = read();
56             add(x1, y1, z);
57             add(x1, y2 + 1, -z);
58             add(x2 + 1, y1, -z);
59             add(x2 + 1, y2 + 1, z);
60         }
61         else printf("%d\n", query(x2, y2) - query(x1 - 1, y2) - query(x2, y1 - 1) + query(x1 - 1, y1 - 1));
62     }
63     return 0;
64 }

AC Code

 

标签：洛谷,limits,int,sum,ret,七分钟,数组,造题,include
来源： https://www.cnblogs.com/shl-blog/p/11285222.html