2020 cattle off winter training camp algorithm base 6 F Cross Array

https://ac.nowcoder.com/acm/contest/3007/F
Title Description

Q learned little new kind of magic, can cause damage to the enemy on the grid of N rows and M columns

The i-th magic can cross a region on the grid (i.e., xi and the first row and the second column yi) causing damage to the enemy zi for each of the squares in

Q small case now using a total of H times magic, you need to count all of the damage after casting is completed, as detailed output description

Reminder: the larger the scale of this problem input, use efficient input mode

1≤H≤500,000 1≤xi, yi, zi, N, M≤2000 1≤xi≤N, 1≤yi≤M

Thinking 

 First, for each coordinate , in which it ranks are affected, so to a certain point , the end result is that the total impact of the overall impact of the line where he received + where the column received, note that if the middle once this point because the impact is born, it will be more calculated once , so with the impact of record for each point by using two arrays, respectively, to maintain the tree rows and columns, when receiving the impact of the tree in an array of rows and the columns are coupled with the rear impact, then , deduplication affect the current point, where the last seek influence the row and column for each point, and then (the current point) is to affect the summation point.

Multiplied by the sum to a modulo

#include <bits/stdc++.h>
#define mem(a, b) memset(a, b, sizeof a)
using namespace std;
const int N = 2010;
typedef long long ll;
const int mod = 1e9 + 7;
template <class T>
bool read(T & a){
	a = 0;
	int flag = 0;
	char ch;
	if((ch = getchar()) == '-'){
		flag = 1;
	}
	else if(ch >= '0' && ch <= '9'){
		a = a * 10 + ch - '0';
	}
	while ((ch = getchar()) >= '0' && ch <= '9'){
		a = a * 10 + ch - '0';
	}
	if(flag)a = -a;
	return true;
}
template <class T, class ... R>
bool read(T & a, R & ... b){
	if(!read(a))return 0;
	read(b...);
}
template <class T>
bool out(T a){
	if(a < 0)putchar('-');
	if(a >= 10)out(a / 10);
	putchar(a % 10 + '0');
	return true;
}
template <class T, class ... R>
bool out(T a, R ... b){
	if(!out(a))return 0;
	out(b...);
}
ll n, m, h;
ll w[N][N];
ll a[N][2];
ll lowbit(ll x){
	return x & -x;
}
ll add(ll x, ll val, ll ind){
	if (ind == 0){
		while (x <= n){
			// ά»¤ÐÐ
			a[x][0] += val;
			x += lowbit(x); 
		}
	}
	else {
		while (x <= m){
			a[x][1] += val;
			x += lowbit(x);
		}
	}
}
ll query(ll x, ll ind){
	ll res = 0;
	while (x){
		res += a[x][ind];
		x -= lowbit(x);
	}
	return res;
}
int main()
{
	read(n, m, h);
	mem(w, 0);
	mem(a, 0);
	for (int i = 1; i <= h; i++){
		ll x, y, z;
		read(x, y, z);
		add(x, z, 0);
		add(y, z, 1);
		w[x][y] -= z;
	}
	ll res = 0;
	for (int i = 1; i <= n; i++){
		for (int j = 1; j <= m; j++){
			ll t1 = query(i, 0) - query(i - 1, 0);
			ll t2 = query(j, 1) - query(j - 1, 1);
			w[i][j] += t1 + t2;
			res += w[i][j] % mod * (i + j) % mod;
			res %= mod;
		}
	}
	out(res % mod);
	return 0;
}