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

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[][] ;

 int n,m,q;

 int a,x,y;

 int l[],h[];

 int ladd[],hadd[];

 int main()

 {

     scanf("%d",&t);

     for(int i=;i<=t;i++)

     {

         scanf("%d %d %d",&n,&m,&q);

         for(int j=;j<=n;j++)

          for(int k=;k<=m;k++)

           scanf("%d",&mp[j][k]);

           for(int j=;j<=n;j++)

           {

           h[j]=j;hadd[j]=j;

           }

           for(int j=;j<=m;j++)

           {

           l[j]=j; ladd[j]=;

           }

           memset(hadd,,sizeof(hadd));

           memset(ladd,,sizeof(ladd));

           int t;

           for(int j=;j<=q;j++)

           {

                scanf("%d %d %d",&a,&x,&y);

                if(a==)

                {

                    t=h[y];

                 h[y]=h[x];

                    h[x]=t;

              }

                else

                if(a==)

                {

                    t=l[y];

                 l[y]=l[x];

                    l[x]=t;

                }

                else

                if(a==)

                {

                    hadd[h[x]]+=y;

               }

                else

                   ladd[l[x]]+=y;

           }

           for(int j=;j<=n;j++)

            {

                printf("%d",mp[h[j]][l[]]+hadd[h[j]]+ladd[l[]]);

              for(int k=;k<=m;k++)

            {

                    printf(" %d",mp[h[j]][l[k]]+hadd[h[j]]+ladd[l[k]]);

            }

            printf("\n");

            }

     }

 return ;

 }

巴特西

HDU 5671 矩阵

Matrix

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