hdoj 5671 Matrix (模拟+技巧)

Matrix

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1392    Accepted Submission(s): 554

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 matrix M 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 
 

 

按着题意去模拟，RE了，看了大神博客，技巧用的挺好的，
数组模拟矩阵每个位置变化和数值变化 ，最后按变化后的行列坐标和变化的值输出

注意行列变换的时候，也要进行变化值的交换

代码：

/*技巧，数组模拟矩阵每个位置变化和数值变化 */#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[1010][1010]; int main() { int t,n,m,q; int s,x,y; int row[1010],col[1010]; int sumr[1010],sumc[1010]; scanf("%d",&t); while(t--) { scanf("%d%d%d",&n,&m,&q); int i,j; for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { scanf("%d",&a[i][j]); } } memset(sumr,0,sizeof(sumr)); memset(sumc,0,sizeof(sumc)); for(i=1;i<=n;i++)row[i]=i; for(j=1;j<=m;j++)col[j]=j; for(i=0;i<q;i++) { scanf("%d%d%d",&s,&x,&y); if(s==1)     {swap(row[x],row[y]);swap(sumr[x],sumr[y]); } if(s==2) { swap(col[x],col[y]); swap(sumc[x],sumc[y]);} if(s==3) sumr[x]+=y; if(s==4) sumc[x]+=y; } for(i=1;i<=n;i++) { for(j=1;j<m;j++) printf("%d ",a[row[i]][col[j]]+sumr[i]+sumc[j]); printf("%d\n",a[row[i]][col[m]]+sumr[i]+sumc[m]); } } return 0; }