HDOJ 5671 Matrix

Matrix

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

Total Submission(s): 780    Accepted Submission(s): 330

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 integerT(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 matrixM after all q operations.

 

Sample Input

23 4 21 2 3 42 3 4 53 4 5 61 1 23 1 102 2 21 1010 11 1 22 1 2

 

Sample Output

12 13 14 151 2 3 43 4 5 61 1010 1
Hint
 Recommand to use scanf and printf 
 

 

中文题面：BC Round #81 (div.2)1002  

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int map[1010][1010];int row[1010],col[1010];int hx[1010],hy[1010]; int main(){int t,n,m,q,i,j,a,x,y;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&q);for(i=1;i<=n;++i){for(j=1;j<=m;++j){scanf("%d",&map[i][j]);row[i]=i; hx[i]=0;}}for(j=1;j<=m;++j){col[j]=j;hy[j]=0;}while(q--){scanf("%d%d%d",&a,&x,&y);if(a==1)swap(row[x],row[y]);else if(a==2)swap(col[x],col[y]);else if(a==3)hx[row[x]]+=y;else hy[col[x]]+=y; }for(i=1;i<=n;++i){for(j=1;j<=m;++j){if(j==m)printf("%d\n",map[row[i]][col[j]]+hx[row[i]]+hy[col[j]]);elseprintf("%d ",map[row[i]][col[j]]+hx[row[i]]+hy[col[j]]);}} } return 0;}