luogu P4514ybtoj树状数组课堂过关差分 例题6区间修改区间查询 & 上帝造题的七分钟

Posted 2021-06-19 SSL_ZZL

【例题6】区间修改区间查询 & 上帝造题的七分钟

• Link
• 题目大意
• 解题思路
• Code

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Link

luogu P4514 上帝造题的七分钟
ybtoj 【树状数组课堂过关】 【例题6】区间修改区间查询
题面//因为不知道侵不侵权所以就是题面是私密的，有账号的直接看转送门就可了

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题目大意

给一个矩阵A，有两个操作

• 1 a, b, c, d, s=》$A_{(a到c),(b到d)}$+s
• 2 a, b, c, d =》求$A_{(a到c),(b到d)}$的和

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解题思路

对，捡了道紫题的便宜

很明显，这种矩阵（区间）加值要用差分
【例题4】区间修改区间查询 这是道一维的差分题，但是这题是二维的

设a[][]是实际数组，s[][]是差分数组
$$a[i][j]=\sum _{k=1}^i\sum _{l=1}^{j}s[k][l]$$

差分时记得要容斥

add(a, b, s); 
add(a, d + 1, -s); 
add(c + 1, b, -s); 
add(c + 1, d + 1, s); 

设f[i][j]是矩阵$A_{(1到i),(1到j)}$的和，那么
$$f[i][j]=\sum _{ia=1}^n\sum _{ja=1}^{m}a[ia][ja]=\sum _{is=1}^n\sum _{js=1}^m\sum _{k=1}^{ia}\sum _{l=1}^{ja}s[k][l]$$

和一维一样，考虑每个s[][]的出现次数
有点复杂，是不会讲的，所以。。。推篇博客？

ybtoj和luogu的输入不太一样

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Code

ybtoj版

#include <iostream>
#include <cstdio>
#define ll long long

using namespace std;

int n, m, a, b, c, d, now;
ll s, tree_s[3000][3000], tree_i[3000][3000], tree_j[3000][3000], tree_ij[3000][3000];

int lowbit(int x) {return (x & -x);}

void add(int x, int y, ll s) {
	for(int i = x; i <= n; i += lowbit(i))
		for(int j = y; j <= m; j += lowbit(j)) {
			tree_s[i][j] += s;
			tree_i[i][j] += s * x;
			tree_j[i][j] += s * y;
			tree_ij[i][j] += s * x * y;
		}
}

ll sum(int x, int y) {
	ll ans = 0;
	for(int i = x; i; i -= lowbit(i))
		for(int j = y; j; j -= lowbit(j))
			ans += tree_s[i][j] * (x + 1) * (y + 1) - tree_i[i][j] * (y + 1) - tree_j[i][j] * (x + 1) + tree_ij[i][j];
	return ans;
}

int main() {
	scanf("%d %d", &n, &m);
	while(scanf("%d", &now) != EOF) {
		scanf("%d %d %d %d", &a, &b, &c, &d);
		if (now == 1) {
			scanf("%lld", &s);
			add(a, b, s); 
			add(a, d + 1, -s); 
			add(c + 1, b, -s); 
			add(c + 1, d + 1, s); 
		} else {
			printf("%lld\\n", sum(c, d) - sum(a - 1, d) - sum(c, b - 1) + sum(a - 1, b - 1)); 
		}
	}
}

luogu版

#include <iostream>
#include <cstdio>

using namespace std;

char now;
int n, m, a, b, c, d;
int s, tree_s[3000][3000], tree_i[3000][3000], tree_j[3000][3000], tree_ij[3000][3000];

int lowbit(int x) {return (x & -x);}

void add(int x, int y, int s) {
	for(int i = x; i <= n; i += lowbit(i))
		for(int j = y; j <= m; j += lowbit(j)) {
			tree_s[i][j] += s;
			tree_i[i][j] += s * x;
			tree_j[i][j] += s * y;
			tree_ij[i][j] += s * x * y;
		}
}

int sum(int x, int y) {
	int ans = 0;
	for(int i = x; i; i -= lowbit(i))
		for(int j = y; j; j -= lowbit(j))
			ans += tree_s[i][j] * (x + 1) * (y + 1) - tree_i[i][j] * (y + 1) - tree_j[i][j] * (x + 1) + tree_ij[i][j];
	return ans;
}

int main() {
	scanf("X %d %d", &n, &m);
	while(~scanf("%s", &now)) {
		scanf("%d %d %d %d", &a, &b, &c, &d);
		if (now == 'L') {
			scanf("%d", &s);
			add(a, b, s); 
			add(a, d + 1, -s); 
			add(c + 1, b, -s); 
			add(c + 1, d + 1, s); 
		} else {
			printf("%d\\n", sum(c, d) - sum(a - 1, d) - sum(c, b - 1) + sum(a - 1, b - 1)); 
		}
	}
}