BestCoder Round #81 (div.2)

 

Matrix

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 357    Accepted Submission(s): 158

对于交换行、交换列的操作，分别记录当前状态下每一行、每一列是原始数组的哪一行、哪一列即可。

对每一行、每一列加一个数的操作，也可以两个数组分别记录。注意当交换行、列的同时，也要交换增量数组。

输出时通过索引找到原矩阵中的值，再加上行、列的增量。

复杂度$O(q+mn)$

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)

 

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <algorithm>
#include <set>
#define LL long long
using namespace std;
int A[1010][1010];
int H[1010],L[1010];
int AddH[1010],AddL[1010];
int main()
{
   int n,m,q;
   int u,x,y;
   int t;
   cin>>t;
   while(t--)
   {
       scanf("%d%d%d",&n,&m,&q);
       for(int i=1;i<=n;i++)
        H[i]=i;
       for(int j=1;j<=m;j++)
        L[j]=j;
       memset(AddH,0,sizeof(AddH));
       memset(AddL,0,sizeof(AddL));
       for(int i=1;i<=n;i++)
       {
           for(int j=1;j<=m;j++)
           {
               scanf("%d",&A[i][j]);
           }
       }
       while(q--)
       {
           scanf("%d%d%d",&u,&x,&y);
           if(u==1)
           {
               swap(H[x],H[y]);
               swap(AddH[x],AddH[y]);
           }
           else if(u==2)
           {
                swap(L[x],L[y]);
               swap(AddL[x],AddL[y]);
           }
           else if(u==3)
           {
               AddH[x]+=y;
           }
           else
           {
               AddL[x]+=y;
           }
       }
       for(int i=1;i<=n;i++)
       {
           for(int j=1;j<=m;j++)
           {
               if(j!=1)
                printf(" ");
               printf("%d",A[H[i]][L[j]]+AddH[i]+AddL[j]);
           }
           printf("\n");
       }
   }

    return 0;
}