LIS POJ 2533 Longest Ordered Subsequence

 

题目传送门

 1 /*  2  LIS(Longest Increasing Subsequence)裸题：  3  状态转移方程：dp[i] = max (do[i], dp[j] + 1) (1 <= j < i)  4  附带有print输出路径函数  5 */  6 #include <cstdio>  7 #include <algorithm>  8 #include <cstring>  9 using namespace std; 10 11 const int MAXN = 1e3 + 10; 12 int a[MAXN]; 13 int dp[MAXN]; 14 int rt[MAXN]; 15 16 void print(int id, int n) 17 { 18 if (rt[id] != -1) 19  { 20  print (rt[id], n); 21  } 22 printf ("%d%c", a[id], (id == n) ? '\n' : ' '); 23 } 24 25 int main(void) //POJ 2533 Longest Ordered Subsequence 26 { 27 //freopen ("LIS.in", "r", stdin); 28 29 int n; 30 scanf ("%d", &n); 31 for (int i=1; i<=n; ++i) 32  { 33 scanf ("%d", &a[i]); 34  } 35 36 int res = 0; 37 for (int i=1; i<=n; ++i) 38  { 39 dp[i] = 1; rt[i] = -1; 40 for (int j=1; j<i; ++j) 41  { 42 if (a[j] < a[i]) 43  { 44 if (dp[i] < dp[j] + 1) 45  { 46 dp[i] = dp[j] + 1; 47 rt[i] = j; 48  } 49  } 50  } 51 res = max (res, dp[i]); 52  } 53 54 printf ("%d\n", res); 55 //print (n, n); 56 57 return 0; 58 }