Codeforces #264 (Div. 2) D. Gargari and Permutations

Gargari got bored to play with the bishops and now, after solving the problem about them, he is trying to do math homework. In a math book he have foundk permutations. Each of them consists of numbers1, 2, ..., n in some order. Now he should find the length of the longest common subsequence of these permutations. Can you help Gargari?

You can read about longest common subsequence there: https://en.wikipedia.org/wiki/Longest_common_subsequence_problem

Input

The first line contains two integers n andk(1 ≤ n ≤ 1000; 2 ≤ k ≤ 5). Each of the nextk lines contains integers1, 2, ..., n in some order — description of the current permutation.

Output

Print the length of the longest common subsequence.

Sample test(s)

Input

4 31 4 2 34 1 2 31 2 4 3

Output

3

Note

The answer for the first test sample is subsequence [1, 2, 3].

题意：求k个长度为n的最长公共子序列

思路1：保存每个数在各自串的位置，因为结果是第1个串中的某个可能，所以我们枚举第1个串的可能，然后检查如果一个以a[j]为结束的最长公共子序列成立的情况是，对于每个串的a[i]都在a[j]的前面，那么就有dp[j] = max(dp[j], dp[i]+1)

[cpp] view plaincopyprint?

1. #include <iostream> 
2. #include <cstring> 
3. #include <cstdio> 
4. #include <algorithm> 
5. using namespace std; 
6. const int maxn = 1010; 
7.  
8. int n, k; 
9. int a[maxn][maxn], b[maxn][maxn], dp[maxn]; 
10.  
11. int check(int x,int y) { 
12.     for (int i = 2; i <= k; i++) 
13.         if (b[i][x] > b[i][y]) 
14.             return 0; 
15.     return 1; 
16. } 
17.  
18. int main() { 
19.     scanf("%d%d", &n, &k); 
20.     for (int i = 1; i <= k; i++) 
21.         for (int j = 1; j <= n; j++) { 
22.             scanf("%d", &a[i][j]); 
23.             b[i][a[i][j]] = j; 
24.         } 
25.  
26.     for (int i = 1; i <= n; i++) 
27.         dp[i] = 1; 
28.  
29.     int ans = 0; 
30.     for (int i = 1; i <= n; i++) { 
31.         for (int j = i+1; j <= n; j++) { 
32.             if (check(a[1][i], a[1][j])) 
33.                 dp[j] = max(dp[i]+1, dp[j]); 
34.         } 
35.     } 
36.  
37.     for (int i = 1; i <= n; i++) 
38.         ans = max(ans, dp[i]); 
39.     printf("%d\n", ans); 
40.     return 0; 
41. } 

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 1010;int n, k;int a[maxn][maxn], b[maxn][maxn], dp[maxn];int check(int x, int y) {for (int i = 2; i <= k; i++)if (b[i][x] > b[i][y])return 0;return 1;}int main() {scanf("%d%d", &n, &k);for (int i = 1; i <= k; i++)for (int j = 1; j <= n; j++) {scanf("%d", &a[i][j]);b[i][a[i][j]] = j;}for (int i = 1; i <= n; i++)dp[i] = 1;int ans = 0;for (int i = 1; i <= n; i++) {for (int j = i+1; j <= n; j++) {if (check(a[1][i], a[1][j]))dp[j] = max(dp[i]+1, dp[j]);}}for (int i = 1; i <= n; i++)ans = max(ans, dp[i]);printf("%d\n", ans);return 0;}

思路2：如果一个数字i在每个串的位置都在j前面，那么i到j就有一条有向边，那么题目就转换为DAG求最长

[cpp] view plaincopyprint?

1. #include <iostream> 
2. #include <cstdio> 
3. #include <cstring> 
4. #include <algorithm> 
5. #include <vector> 
6. using namespace std; 
7. const int maxn = 1010; 
8.  
9. int num[10][maxn], vis[maxn]; 
10. int n, k; 
11. vector<int> g[maxn]; 
12.  
13. int check(int x,int y) { 
14.     for (int i = 0; i < k; i++) 
15.         if (num[i][x] >= num[i][y]) 
16.             return 0; 
17.     return 1; 
18. } 
19.  
20. int dfs(int x) { 
21.     int ans = 0; 
22.     if (vis[x]) 
23.         return vis[x]; 
24.  
25.     int size = g[x].size(); 
26.     for (int i = 0; i < size; i++) 
27.         ans = max(ans, dfs(g[x][i])); 
28.  
29.     return vis[x] = ans + 1; 
30. } 
31.  
32. int main() { 
33.     scanf("%d%d", &n, &k); 
34.     memset(vis, 0, sizeof(vis)); 
35.     for (int i = 0; i <= n; i++) 
36.         g[i].clear(); 
37.  
38.     int a; 
39.     for (int i = 0; i < k; i++) 
40.         for (int j = 1; j <= n; j++) { 
41.             scanf("%d", &a); 
42.             num[i][a] = j; 
43.         } 
44.  
45.     for (int i = 1; i <= n; i++)  
46.         for (int j = 1; j <= n; j++)  
47.             if (check(i, j)) 
48.                 g[i].push_back(j); 
49.      
50.     int ans = 0; 
51.     for (int i = 1; i <= n; i++) 
52.         if (!vis[i]) 
53.             ans = max(ans, dfs(i)); 
54.  
55.     printf("%d\n", ans); 
56.     return 0; 
57. }