AcWing 798. 差分矩阵

Posted 2022-02-09 MangataTS

题目连接

https://www.acwing.com/problem/content/800/

思路

差分矩阵注意的是在构造差分矩阵的时候不是一维的和前一个数相减，而是看作一个小矩阵进行操作

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000009
#define endl "\\n"
#define PII pair<int,int>
ll ksm(ll a,ll b) 
	ll ans = 1;
	for(;b;b>>=1LL) 
		if(b & 1) ans = ans * a % mod;
		a = a * a % mod;
	
	return ans;


ll lowbit(ll x)return -x & x;

const int N = 1e3+10;
int n,m,q,a[N][N],d[N][N];

void insert(int x1,int y1,int x2,int y2,int c)
	d[x1][y1] += c;
	d[x2 + 1][y1] -= c;
	d[x1][y2 + 1] -= c;
	d[x2 + 1][y2 + 1] += c;


int main()

	cin>>n>>m>>q;
	for(int i = 1;i <= n; ++i) 
		for(int j = 1;j <= m; ++j) 
			cin>>a[i][j];
		
	
	for(int i = 1;i <= n; ++i) 
		for(int j = 1;j <= m; ++j) 
			insert(i,j,i,j,a[i][j]);
		
	
	int x1,y1,x2,y2,c;
	while(q--) 
		cin>>x1>>y1>>x2>>y2>>c;
		insert(x1,y1,x2,y2,c);
	
	ll res = 0;
	for(int i = 1;i <= n; ++i) 
		for(int j = 1;j <= m; ++j) 
			d[i][j] += d[i-1][j] + d[i][j-1] - d[i-1][j-1];//求前缀和，需要用到之前求出来的状态
			cout<<d[i][j]<<" \\n"[j==m];
		
	
	
	return 0;