Matrix

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 502    Accepted Submission(s): 215

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≤M,i,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

 

Recommend

wange2014   

题目大意：有一个n行m列的矩阵，在这个矩阵上进行q个操作，输出经过所有q个操作以后的矩阵M

思路：用4个数组，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 
      
       #include 
       
         using namespace std;int h[1010],l[1010], zh[1010], zl[1010];int a[1010][1010];int main(){    int T, n, m, q, i, j, c, x, y, t;    cin>>T;    while(T--)    {    memset(zh,0,sizeof(zh));    memset(zl,0,sizeof(zl));    for(i=1;i<=1005;i++)    h[i]=l[i]=i;    scanf("%d%d%d", &n, &m, &q);    for(i=1;i<=n;i++)        for(j=1;j<=m;j++)    scanf("%d", &a[i][j]);    while(q--)    {    scanf("%d%d%d", &c, &x, &y);    if(c==1) {t=h[x];h[x]=h[y];h[y]=t;}    else if(c==2) {t=l[x];l[x]=l[y];l[y]=t;}    else if(c==3) zh[h[x]] += y;    else if(c==4) zl[l[x]] += y;    }    for(i=1;i<=n;i++)        {        for(j=1;j<=m;j++)        if(j==1)        printf("%d", a[h[i]][l[j]]+zh[h[i]]+zl[l[j]]);        else        printf(" %d", a[h[i]][l[j]]+zh[h[i]]+zl[l[j]]);        printf("\n");        }    }    return 0;}