hdu 5671 Matrix(模拟)

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
solution：
若直接暴搞的话会T。
对于交换行、交换列的操作，分别记录当前状态下每一行、每一列是原始数组的哪一行、哪一列即可。对每一行、每一列加一个数的操作，也可以两个数组分别记录。注意当交换行、列的同时，也要交换增量数组。输出时通过索引找到原矩阵中的值，再加上行、列的增量。
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 1e3 + 20;long long col[maxn], row[maxn], map[maxn][maxn];int c[maxn], r[maxn];int main(){    int t,x,y,a,n,m,q;    scanf("%d", &t);    while (t--)    {        scanf("%d%d%d", &n, &m, &q);        for (int i = 1; i <= n; i++)        {            r[i] = i; row[i] = 0;            for (int j = 1; j <= m; j++)            {                c[j] = j; col[j] = 0;                scanf("%I64d", &map[i][j]);            }        }        while (q--)        {            scanf("%d%d%d", &a, &x, &y);            switch (a)            {            case 1:swap(r[x], r[y]); swap(row[x], row[y]); break;            case 2:swap(c[x], c[y]); swap(col[x], col[y]); break;            case 3:row[x] += y; break;            case 4:col[x] += y;            }        }        for (int i = 1; i <= n; i++)            for (int j = 1; j <= m; j++)            printf("%I64d%c", map[r[i]][c[j]] + row[i] + col[j], j == m ? '\n' : ' ');    }    return 0;}