题解 | #【模板】二维差分#

解题思路：

• 建立差分二维数组，每一行按照一维差分数组建立$cf[i][j]=v[i][j]-v[i][j-1],(1\le i\le n,1\le j\le m)cf[i][j]\; =\; v[i][j]\; -\; v[i][j-1],(1\; \backslash leq\; i\; \backslash leq\; n,1\; \backslash leq\; j\; \backslash leq\; m)$
• 对于区间更新q，只要对每行按一维的更新$cf[i][y_1]+=a,cf[i][y_2+1]-=a,(x_1\le i\le x_2)cf[i][y\_1]+=a,\; cf[i][y\_2+1]\; -=a,(x\_1\; \backslash leq\; i\; \backslash leq\; x\_2)$

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n, m, q;
    cin>>n>>m>>q;
    vector<vector<long long>> v(n+1, vector<long long>(m+1, 0));
    vector<vector<long long>> cf(n+1, vector<long long>(m+1, 0));
    for(int i = 1; i <=n; ++i){
        for(int j = 1; j <= m; ++j){
            scanf("%lld", &v[i][j]);
            cf[i][j] = v[i][j] - v[i][j-1];
        }
    }
    int x1, y1, x2, y2, a;
    for(int i = 1; i <= q; ++i){
        scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&a);
        for(int j = x1; j <= x2; ++j){
            cf[j][y1] += a;
            if(y2+1<=m) cf[j][y2+1] -= a;
        }
    }
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <=m; ++j){
            v[i][j] = v[i][j-1]+cf[i][j];
            if (j!= m)cout<<v[i][j]<<" ";
            else cout<<v[i][j]<<endl;
        }
    }
    return 0;
}