Hdu5671 Matrix 矩阵、行列交换+行列加值

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 integer T(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 
   
    
  

Source

BestCoder Round #81 (div.2) 

对一个矩阵进行行列交换以及加值操作，直接按题所给方式去做必然超时。

但考虑操作都是针对整一行，整一列的，可以考虑用数组去记录操作的行列号，而不是直接移动矩阵元素。

include <iostream>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
struct node{
    int s;
    int add;
}row[1005],column[1005];
int a[1005][1005];
int main()
{
    int t,n,m,s,o,x,y;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&m,&s);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%d",&a[i][j]);
            }
        }
        for(int i=1;i<=n;i++){
            row[i].add=0;
            row[i].s=i;
        }
        for(int i=1;i<=m;i++){
            column[i].add=0;
            column[i].s=i;
        }
        for(int i=0;i<s;i++){
            scanf("%d%d%d",&o,&x,&y);
            if(o==1){
                swap(row[x],row[y]);
            }else if(o==2){
                swap(column[x],column[y]);
            }else if(o==3){
                row[x].add+=y;
            }else{
                column[x].add+=y;
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                printf(j==1? "%d":" %d",a[row[i].s][column[j].s]+row[i].add+column[j].add);
            }
            printf("\n");
        }
    }
    return 0;
}