【二维差分】 差分矩阵

传送门

题意

给定一个\(n\times m\)的初始矩阵，\(q\)个操作，每个操作包含\(x_{1},y_{1},x_{2},y_{2},c\)，使\((x1, y1)\)为左上角，\((x2, y2)\)为右下角的子矩阵中的所有元素加上\(c\)
求所有操作后的矩阵

数据范围

\(\begin{array}{l}1 \leq n, m \leq 1000 \\ 1 \leq q \leq 100000 \\ 1 \leq x 1 \leq x 2 \leq n \\ 1 \leq y 1 \leq y 2 \leq m \\ -1000 \leq c \leq 1000 \\ -1000 \leq \text { 矩阵内元素的值 } \leq 1000\end{array}\)

题解

\(S[x1, y1] \:+= c\)
\(S[x2 + 1, y1] \:-= c\)
\(S[x1, y2 + 1] \:-= c\)
\(S[x2 + 1, y2 + 1] \:+= c\)

Code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e3+10;
int a[N][N],b[N][N];
int n,m,q;
inline void insert(int x1,int y1,int x2,int y2,int c){
    b[x1][y1]+= c;
    b[x2+1][y1]-=c;
    b[x1][y2+1]-=c;
    b[x2+1][y2+1]+=c;
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin>>n>>m>>q;
    b[0][0]=0;
    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]);
        while(q--){
            int x1,y1,x2,y2,c;
            cin>>x1>>y1>>x2>>y2>>c;
            insert(x1,y1,x2,y2,c);
        }

        for(int i=1;i<=n;i++) {
            for (int j = 1; j <= m; j++) {
                b[i][j] = b[i][j] + b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];
                cout << b[i][j] << ' ';
            }
            cout<<endl;
        }
}