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 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 

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]]);
				else
					printf("%d ",map[row[i]][col[j]]+hx[row[i]]+hy[col[j]]);
			}
		}
	}
	return 0;
}