HDOJ 5671-Matrix【矩阵的变换】

骑猪看日落 2022-07-24 07:17 177阅读 0赞

# Matrix #

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

Total Submission(s): 1049 Accepted Submission(s): 434

Problem Description

There is a matrix  M![Image 1][] that has  n![Image 1][] rows and  m![Image 1][] columns  (1≤n≤1000,1≤m≤1000)![Image 1][].Then we perform  q(1≤q≤100,000)![Image 1][] operations:  
  
1 x y: Swap row x and row y  (1≤x,y≤n)![Image 1][];  
  
2 x y: Swap column x and column y  (1≤x,y≤m)![Image 1][];  
  
3 x y: Add y to all elements in row x  (1≤x≤n,1≤y≤10,000)![Image 1][];  
  
4 x y: Add y to all elements in column x  (1≤x≤m,1≤y≤10,000)![Image 1][];

Input

There are multiple test cases. The first line of input contains an integer  T(1≤T≤20)![Image 1][] indicating the number of test cases. For each test case:  
  
The first line contains three integers  n![Image 1][],  m![Image 1][] and  q![Image 1][].  
The following  n![Image 1][] lines describe the matrix M.  (1≤M![Image 1][]i,j![Image 1][]≤10,000)![Image 1][] for all  (1≤i≤n,1≤j≤m)![Image 1][].  
The following  q![Image 1][] lines contains three integers  a(1≤a≤4)![Image 1][],  x![Image 1][] and  y![Image 1][].

Output

For each test case, output the matrix  M![Image 1][] after all  q![Image 1][] 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) ][BestCoder Round _81 _div.2_]

解题思路：

题目大意：

有一个nn行mm列的矩阵(1 \leq n \leq 1000 ,1 \leq m \leq 1000 )(1≤n≤1000,1≤m≤1000)，在这个矩阵上进行qq  (1 \leq q \leq 100,000)(1≤q≤100,000) 个操作：
    
    1 x y: 交换矩阵MM的第xx行和第yy行(1 \leq x,y \leq n)(1≤x,y≤n);
    2 x y: 交换矩阵MM的第xx列和第yy列(1 \leq x,y \leq m)(1≤x,y≤m);
    3 x y: 对矩阵MM的第xx行的每一个数加上y(1 \leq x \leq n,1 \leq y \leq 10,000)y(1≤x≤n,1≤y≤10,000);
    4 x y: 对矩阵MM的第xx列的每一个数加上y(1 \leq x \leq m,1 \leq y \leq 10,000)y(1≤x≤m,1≤y≤10,000);

对于矩阵来说，我们把他输出的坐标顺序记录下就行了。行和列都要记录。

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    #define LL long long
    using namespace std;
    struct node
    {
    	LL wei;
    	LL bei;
    }num[1005];
    struct node2
    {
    	LL lie;
    	LL bb;
    }numline[1005];
    LL map[1005][1005];
    void init(LL n,LL m)
    {
    	for(LL i=1;i<=n;i++)
    	{
    		num[i].wei=i;
    		num[i].bei=0;
    	}
    	for(LL i=1;i<=m;i++)
    	{
    		numline[i].lie=i;
    		numline[i].bb=0;
    	}
    }
    int main()
    {
    	int t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		LL n,m,p;
    		scanf("%lld%lld%lld",&n,&m,&p);
    		init(n,m);
    		for(LL i=1;i<=n;i++)
    		{
    			for(LL j=1;j<=m;j++)
    			{
    				scanf("%lld",&map[i][j]);
    			}
    		}
    		while(p--)
    		{
    			LL biao,xx,yy;
    			scanf("%lld%lld%lld",&biao,&xx,&yy);
    			if(biao==1)
    			{
    				node tt;
    				tt=num[xx];
    				num[xx]=num[yy];
    				num[yy]=tt;
    			}
    			else if(biao==2)
    			{
    				node2 tt;
    				tt=numline[xx];
    				numline[xx]=numline[yy];
    				numline[yy]=tt;
    			}
    			else if(biao==3)
    			{
    				num[xx].bei+=yy;
    			}
    			else
    			{
    				numline[xx].bb+=yy;
    			}
    		}
    		for(LL i=1;i<=n;i++)
    		{
    			for(LL j=1;j<=m;j++)
    			{
    				if(j==1)
    				printf("%lld",map[num[i].wei][numline[j].lie]+num[i].bei+numline[j].bb);
    				else
    				printf(" %lld",map[num[i].wei][numline[j].lie]+num[i].bei+numline[j].bb);
    			}
    			printf("\n");
    		}
    	}
    	return 0;
     }

[Image 1]: 
[BestCoder Round _81 _div.2_]: http://acm.hdu.edu.cn/search.php?field=problem&key=BestCoder+Round+%2381+%28div.2%29&source=1&searchmode=source