HDU 5671 矩阵

Matrix

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 271    Accepted Submission(s): 126

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 42

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)

 

题意：对矩阵执行q次  4种类型的操作 输出 最终矩阵

 

题解：

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

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #include<algorithm>
 #define LL __int64
 #define pi acos(-1.0)
 #define mod 1
 #define maxn 10000
 using namespace std;
int t;
int mp[1005][1005] ;
int n,m,q;
int a,x,y;
int l[1005],h[1005];
int ladd[1005],hadd[1005];
int main()
{
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        scanf("%d %d %d",&n,&m,&q);
        for(int j=1;j<=n;j++)
         for(int k=1;k<=m;k++)
          scanf("%d",&mp[j][k]);
          for(int j=1;j<=n;j++)
          {
          h[j]=j;hadd[j]=j;
          }
          for(int j=1;j<=m;j++)
          {
          l[j]=j; ladd[j]=0;
          }
          memset(hadd,0,sizeof(hadd));
          memset(ladd,0,sizeof(ladd));
          int t;
          for(int j=1;j<=q;j++)
          {
               scanf("%d %d %d",&a,&x,&y);
               if(a==1)
               {
                   t=h[y];
                h[y]=h[x];
                   h[x]=t;
             }
               else
               if(a==2)
               {
                   t=l[y];
                l[y]=l[x];
                   l[x]=t;
               }
               else
               if(a==3)
               {
                   hadd[h[x]]+=y;
              }
               else
                  ladd[l[x]]+=y;
          }
          for(int j=1;j<=n;j++)
           {
               printf("%d",mp[h[j]][l[1]]+hadd[h[j]]+ladd[l[1]]);
             for(int k=2;k<=m;k++)
           {
                   printf(" %d",mp[h[j]][l[k]]+hadd[h[j]]+ladd[l[k]]);
           }
           printf("
");
           }
    }
return 0; 
}