Codeforces 390E Inna and Large Sweet Matrix 树状数组改段求段

题目链接：点击打开链接

题意：给定n*m的二维平面 w个操作

int mp[n][m] = { 0 };

1、0 (x1,y1) (x2,y2) value

for i : x1 to x2

for j : y1 to y2

mp[i][j] += value;

2、1 (x1, y1) (x2 y2)

ans1 = 纵坐标在 y1,y2间的总数

ans2 = 横坐标不在x1,x2间的总数

puts(ans1-ans2);

more format:

for i : 1 to n

for j : y1 to y2

ans1 += mp[i][j]

由于n最大是4e6, 所以用树状数组改段求段取代线段树

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
template <class T>
inline bool rd(T &ret) {
	char c; int sgn;
	if (c = getchar(), c == EOF) return 0;
	while (c != '-' && (c<'0' || c>'9')) c = getchar();
	sgn = (c == '-') ? -1 : 1;
	ret = (c == '-') ? 0 : (c - '0');
	while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
	ret *= sgn;
	return 1;
}
template <class T>
inline void pt(T x) {
	if (x <0) {
		putchar('-');
		x = -x;
	}
	if (x>9) pt(x / 10);
	putchar(x % 10 + '0');
}
using namespace std;
typedef long long ll;
const int N = 4e6 + 100;
template<class T>
struct Tree{
	T c[2][N];
	int maxn;
	void init(int x){
		maxn = x+10; memset(c, 0, sizeof c);
	}
	inline int lowbit(int x){ return x&-x; }
	T sum(T *b, int x){
		T ans = 0;
		if (x == 0)ans = b[0];
		while (x)ans += b[x], x -= lowbit(x);
		return ans;
	}
	void change(T *b, int x, T value){
		if (x == 0)b[x] += value, x++;
		while (x <= maxn)b[x] += value, x += lowbit(x);
	}
	T get_pre(int r){
		return sum(c[0], r) * r + sum(c[1], r);
	}
	void add(int l, int r, T value){
		change(c[0], l, value);
		change(c[0], r + 1, -value);
		change(c[1], l, value * (-l + 1));
		change(c[1], r + 1, value * r);
	}
	T get(int l, int r){
		return get_pre(r) - get_pre(l - 1);
	}
};
Tree<ll> x, y;
int main(){
	int n, m, w;
	rd(n); rd(m); rd(w);
	x.init(n); y.init(m);
	ll all = 0;
	while (w--){
		int op, x1, x2, y1, y2; ll value;
		rd(op); rd(x1); rd(y1); rd(x2); rd(y2);
		if (op == 0)
		{
			rd(value);
			all += value * (x2 - x1 + 1) * (y2 - y1 + 1);
			x.add(x1, x2, value * (y2 - y1 + 1));
			y.add(y1, y2, value * (x2 - x1 + 1));
		}
		else
		{
			pt(y.get(1, y2) - y.get(1, y1 - 1) - (all - x.get(1, x2) + x.get(1, x1 - 1))); puts("");
		}
	}
	return 0;
}