HDU5671矩阵行列交换

Problem Description

There is a matrix M that has n rows and m columns (1≤n≤1000,1≤m≤1000).Then we perform q(1≤q≤100,000) operations:

1 x y: Swap row x and row y (1≤x,y≤n);

2 x y: Swap column x and column y (1≤x,y≤m);

3 x y: Add y to all elements in row x (1≤x≤n,1≤y≤10,000);

4 x y: Add y to all elements in column x (1≤x≤m,1≤y≤10,000);

 

Input

There are multiple test cases. The first line of input contains an integerT(1≤T≤20) indicating the number of test cases. For each test case:

The first line contains three integers n,m and q.
The following n lines describe the matrix M.(1≤Mi,j≤10,000) for all (1≤i≤n,1≤j≤m).
The following q lines contains three integers a(1≤a≤4),x and y.

 

Output

For each test case, output the matrix M after all q operations.

 

Sample Input

2
3 4 2
1 2 3 4
2 3 4 5
3 4 5 6
1 1 2
3 1 10
2 2 2
1 10
10 1
1 1 2
2 1 2

 

Sample Output

12 13 14 15
1 2 3 4
3 4 5 6
1 10
10 1

Hint
 Recommand to use scanf and printf 

#include<cstdio>
#include<algorithm>
#include<cstring>
typedef long long ll;
using namespace std;
int matrix[1001][1001];
struct node
{
    int pos;//位置
    int num;//数值
};
node row[1001],col[1001];
int main()
{
    int T,n,m,x,y,i,j,messege,q;
    node t;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&q);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            scanf("%d",&matrix[i][j]);
        }
        for(i=1;i<=n;i++)
        {
            row[i].pos=i;
            row[i].num=0;
        }
        for(i=1;i<=m;i++)
        {
            col[i].pos=i;
            col[i].num=0;
        }
        while(q--)
        {
            scanf("%d%d%d",&messege,&x,&y);
            switch(messege)
            {
                case 1://记录当前行的数据
                    t=row[x];
                    row[x]=row[y];
                    row[y]=t;
                    break;
                case 2://记录当前列的数据
                    t=col[x];
                    col[x]=col[y];
                    col[y]=t;
                    break;
                case 3:
                    row[x].num+=y;
                    break;
                case 4:
                    col[x].num+=y;
                    break;
            }
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(j!=m)
                printf("%d ",matrix[row[i].pos][col[j].pos]+row[i].num+col[j].num);
                else
                printf("%d",matrix[row[i].pos][col[j].pos]+row[i].num+col[j].num);
            }
            printf("\n");
        }
    }
    return 0;
}