P4514 上帝造题的七分钟
题意: 二维区间修改 区间查询
**错误日志: 写了个 4 重循环忘记调用 (i) **
Solution
二维树状数组 巨尼玛毒瘤
听说二维线段树会 (MLE) 反正我两个都不会 就学了二维树状数组
然而这是什么东西啊太恶心了
首先是推导:
设 (b[x][y]) 为二维增量数组, 那么有:
[SUM[x,y]=sum_{i = 1}^{x}sum_{j = 1}^{y}sum_{k = 1}^{i}sum_{l = 1}^{j}b[k][l]$$ $$=sum_{i = 1}^{x}sum_{j = 1}^{y}(x - i + 1)(y - j + 1)b[i][j]$$把有关 $i, j$ 的丢到一边, 展开化简得
$$(xy + x + y + 1)sum_{i = 1}^{x}sum_{j = 1}^{y}b[i][j] - (x +1)sum_{i = 1}^{x}sum_{j = 1}^{y}j * b[i][j] - (y + 1)sum_{i = 1}^{x}sum_{j = 1}^{y}i * b[i][j] + sum_{i = 1}^{x}sum_{j = 1}^{y}ij * b[i][j]]
所以, 我们维护 (b[i][j], b[i][j] * i, b[i][j] * j, b[i][j] * i * j) 这四个前缀和即可 二维树状数组区间修改区间查询
Code
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
using namespace std;
int RD(){
int out = 0,flag = 1;char c = getchar();
while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
return flag * out;
}
const int maxn = 4019;
int lenx, leny;
int num, na;
#define lowbit(i) ((i) & (-i))
int c[maxn][maxn][4];
void update(int x, int y, int val, int o){
if(x < 1 || x > lenx || y < 1 || y > leny)return ;
if(o == 0)val = val;
else if(o == 1)val *= x;
else if(o == 2)val *= y;
else val *= x * y;
for(int i = x;i <= lenx;i += lowbit(i))
for(int j = y;j <= leny;j += lowbit(j))
c[i][j][o] += val;
}
int get_sum(int x, int y, int o){
int ans = 0;
for(int i = x;i > 0;i -= lowbit(i))
for(int j = y;j > 0;j -= lowbit(j))
ans += c[i][j][o];
return ans;
}
void uprange(int x1, int y1, int x2, int y2, int val){
for(int i = 0;i < 4;i++){
update(x1, y1, val, i), update(x1, y2 + 1, -val, i);
update(x2 + 1, y2 + 1, val, i), update(x2 + 1, y1, -val, i);
}
}
int query(int x, int y){
return
(x * y + x + y + 1) * get_sum(x, y, 0) -
(x + 1) * get_sum(x, y, 2) -
(y + 1) * get_sum(x, y, 1) +
get_sum(x, y, 3);
}
char cmd;
int main(){
cin>>cmd;lenx = RD();leny = RD();
while(cin>>cmd){
int x1 = RD(), y1 = RD(), x2 = RD(), y2 = RD();
if(cmd == 'L'){
int val = RD();
uprange(x1, y1, x2, y2, val);
}
else{
printf("%d
",
query(x2, y2) +
query(x1 - 1, y1 - 1) -
query(x2, y1 - 1) -
query(x1 - 1, y2)
);
}
}
}