hdu 5671 Matrix(BC——思维题)

题目链接:acm.hdu.edu.cn/showproblem.php?pid=5671

Matrix

Time Limit: 3000/1500 MS (Java/Others)Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 891Accepted Submission(s): 371

Problem Description There is a matrixMthat hasnrows andmcolumns(1≤n≤1000,1≤m≤1000).Then we performq(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 integersn,mandq.
The followingnlines describe the matrix M.(1≤Mi,j≤10,000)for all(1≤i≤n,1≤j≤m).
The followingqlines contains three integersa(1≤a≤4),xandy.

Output

For each test case, output the matrixMafter allqoperations.

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  

Source

BestCoder Round #81 (p.2)

题目大意：

有一个nn行mm列的矩阵(1 \leq n \leq 1000 ,1 \leq m \leq 1000 )(1≤n≤1000,1≤m≤1000)，在这个矩阵上进行qq  (1 \leq q \leq 100,000)(1≤q≤100,000) 个操作：

1 x y: 交换矩阵MM的第xx行和第yy行(1 \leq x,y \leq n)(1≤x,y≤n);
2 x y: 交换矩阵MM的第xx列和第yy列(1 \leq x,y \leq m)(1≤x,y≤m);
3 x y: 对矩阵MM的第xx行的每一个数加上y(1 \leq x \leq n,1 \leq y \leq 10,000)y(1≤x≤n,1≤y≤10,000);
4 x y: 对矩阵MM的第xx列的每一个数加上y(1 \leq x \leq m,1 \leq y \leq 10,000)y(1≤x≤m,1≤y≤10,000);对于每组数据，输出经过所有q个操作之后的矩阵M。

解题思路：如果单纯进行模拟的话，数据量比较大，所以采用标记的方式。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。

详见代码。

#include 
#include 
#include 

using namespace std;

int Map[1010][1010],a;
int h[1010],l[1010],dh[1010],dl[1010];

int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        memset(dh,0,sizeof(dh));
        memset(dl,0,sizeof(dl));
        int n,m,q;
        scanf("%d%d%d",&n,&m,&q);
        for (int i=1; i