【HDU】-5671-Matrix（思维，好）

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

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

看代码理解吧：

#include<cstdio> #include<cstring>#include<algorithm>using namespace std;#define INF 0x3f3f3f3f#define CLR(a,b) memset(a,b,sizeof(a))int a[1010][1010];int row[1010],rad[1010],cul[1010],cad[1010];int main(){int u, n,m,p;scanf("%d",&u);while(u--){CLR(a,0);CLR(row,0),CLR(rad,0);CLR(cul,0),CLR(cad,0);scanf("%d%d%d",&n,&m,&p);for(int i=1;i<=n;i++)row[i]=i,rad[i]=0;for(int i=1;i<=m;i++)cul[i]=i,cad[i]=0;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)scanf("%d",&a[i][j]);int op,x,y;while(p--){scanf("%d%d%d",&op,&x,&y);if(op==1)swap(row[x],row[y]);else if(op==2)swap(cul[x],cul[y]);else if(op==3)rad[row[x]]+=y;else cad[cul[x]]+=y;}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(j>1)printf(" ");printf("%d",a[row[i]][cul[j]]+rad[row[i]]+cad[cul[j]]);}printf("\n");}}return 0;}