codeforces B. Multitasking 解题报告

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int maxn = 1000 + 10;
 8 const int maxm = 100 + 5;
 9 
10 int a[maxn][maxm];
11 int ans[5000][2];
12 
13 int main()
14 {
15     int i, j, l, n, m, k, tj, cnt;
16     while (scanf("%d%d%d", &n, &m, &k) != EOF)
17     {
18         for (i = 1; i <= n; i++)
19         {
20             for (j = 1; j <= m; j++)
21                 scanf("%d", &a[i][j]);
22         }
23         cnt = 0;
24         for (i = 1; i <= n; i++)  // n 行
25         {
26             for (j = 1; j < m; j++)  // m 列
27             {
28                 tj = j;
29                 int tmp = a[i][j];
30                 for (l = j+1; l <= m; l++)   
31                 {
32                     if (tmp > a[i][l] && k == 0 || tmp < a[i][l] && k == 1)
33                     {
34                         tmp = a[i][l];    // 找出第 i 行中第 j 个最少的数
35                         tj = l;
36                     }
37                 }
38                 if (j != tj)
39                 {
40                     swap(a[i][tj], a[i][j]);   // 找完之后要交换
41                     if (k == 0)
42                     {
43                         ans[cnt][0] = j;
44                         ans[cnt][1] = tj;
45                     }
46                     else
47                     {
48                         ans[cnt][0] = tj;
49                         ans[cnt][1] = j;
50                     }
51                     cnt++;
52                     for (l = i+1; l <= n; l++)   // 处理第i+1 ~ n 行的数组
53                     {
54                         if (a[l][j] > a[l][tj] && k == 0 || a[l][j] < a[l][tj] && k == 1)
55                             swap(a[l][j], a[l][tj]);
56                     }
57                 }
58             }       
59         }
60         printf("%d\n", cnt);
61         for (i = 0; i < cnt; i++)
62             printf("%d %d\n", ans[i][0], ans[i][1]); 
63     }
64     return 0;
65 }