HDU 5671 Matrix

Matrix

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1398    Accepted Submission(s): 557

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 
   
    
  

 

Source

BestCoder Round #81 (div.2)

 

Recommend

wange2014

 

这个题好费脑子啊TAT英语很简单，不翻译了。

就是把最先开始的地图的顺序当做二维数组存起来。然后二维数组的角标是不断更新的地图。最后按照角标顺序输出原地图就好。

就是两个图转换这块想起来感觉脑袋都要打弯了= =

#include<stdio.h>
#include<string.h>
int map[1010][1010],x[1010],y[1010];
int addx[1010],addy[1010];
int main()
{
	int T;
	int i,j,a,b,t,n,m,q;
	scanf("%d",&T);
	while(T--)
	{
		memset(addx,0,sizeof(addx));
		memset(addy,0,sizeof(addy));
		scanf("%d%d%d",&n,&m,&q);
		for(i=1;i<=n;i++)
		{
			x[i]=i;
			for(j=1;j<=m;j++)
			{
				y[j]=j;
				scanf("%d",&map[i][j]);
			}
		}
		while(q--)
		{
			scanf("%d%d%d",&t,&a,&b);
			if(t==1)
			{
				int tmp=x[a];
				x[a]=x[b];
				x[b]=tmp;
			}
			else if(t==2)
			{
				int tmp=y[a];
				y[a]=y[b];
				y[b]=tmp;
			}
			else if(t==3)
				addx[x[a]]+=b;
			else
				addy[y[a]]+=b;
		}
		for(i=1;i<=n;i++)
		{
			printf("%d",map[x[i]][y[1]]+addx[x[i]]+addy[y[1]]);
			for(j=2;j<=m;j++)
			{
				printf(" %d",map[x[i]][y[j]]+addx[x[i]]+addy[y[j]]);
			}
			printf("\n");
		}
	}
	return 0;
 }