hdu-5671 Matrix

题目链接：

Matrix

Time Limit: 3000/1500 MS (Java/Others)   

 Memory Limit: 131072/131072 K (Java/Others)

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

 

题意：

 

一个矩阵交换行或者列,或者把某一行或者某一列加一个数;

 

思路：

 

用一个数组记录位置,一个记录加的数值;

这题有个奇妙的TLE点就是我开的的t,x,y为全局变量时会TLE,开成局部的就过了,求懂原理的大神留言讲解;

 

AC代码：

 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const ll inf=1e15;
const int N=1006;
int a[N][N],l[N],r[N],fl[N],fr[N];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,q;
        scanf("%d%d%d",&n,&m,&q);
        for(int i=1;i<=n;i++)l[i]=i,fl[i]=0;
        for(int i=1;i<=m;i++)r[i]=i,fr[i]=0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        int t,x,y;
        for(int i=1;i<=q;i++)
        {
            scanf("%d%d%d",&t,&x,&y);
            if(t==1)
                swap(l[x],l[y]);
            else if(t==2)
                swap(r[x],r[y]);
            else if(t==3)
                fl[l[x]]+=y;
            else
                fr[r[x]]+=y;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<m;j++)
            {
                printf("%d ",a[l[i]][r[j]]+fl[l[i]]+fr[r[j]]);
            }
            printf("%d
",a[l[i]][r[m]]+fl[l[i]]+fr[r[m]]);
        }
    }
    return 0;
}