hdu 5671 Matrix

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

解析：本题很容易超时，需要注意，要用数组来记录数据代替循环，注意理解，代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int s[1005][1005];int p[1005],q[1005];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int g[1005]={0},h[1005]={0};        int a,b,n;        scanf("%d %d %d",&a,&b,&n);        for(int i=1; i<=a; i++)            p[i]=i;        for(int j=1; j<=b; j++)            q[j]=j;        for(int i=1;i<=a;i++)          for(int j=1;j<=b;j++)             scanf("%d",&s[i][j]);        int k,x,y;        while(n--)        {            scanf("%d %d %d",&k,&x,&y);            if(k==1)              swap(p[x],p[y]);    //数组保存的数据对换，代替循环            else if(k==2)              swap(q[x],q[y]);            else  if(k==3)              g[p[x]]+=y;         //利用数组下标的数值加y代替循环，处理方法很巧妙            else              h[q[x]]+=y;        }         for(int i=1;i<=a;i++)        {            for(int j=1;j<b;j++)               printf("%d ",s[p[i]][q[j]]+g[p[i]]+h[q[j]]);            printf("%d\n",s[p[i]][q[b]]+g[p[i]]+h[q[b]]);        }     }    return 0;}