POJ_2533 Longest Ordered Subsequence 【LIS】
一、题目
二、分析
动态规划里的经典问题。重在DP思维。
如果用最原始的DP思想做,状态转移方程为$DP[i] = max(DP[j] + 1)$,$j$满足$j<i$,且$a[i]>a[j]$。
三、AC代码
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5 #include <vector>
6 #include <cmath>
7
8 using namespace std;
9 #define ll long long
10 #define Min(a,b) ((a)>(b)?(b):(a))
11 #define Max(a,b) ((a)>(b)?(a):(b))
12 const int MAXN = 1000;
13 int A[MAXN + 13], Len;
14
15 int main()
16 {
17 int N;
18 while(scanf("%d", &N) != EOF) {
19 int data;
20 Len = 0;
21 for(int i = 0; i < N; i++) {
22 scanf("%d", &data);
23 if(!Len)
24 A[Len++] = data;
25 else {
26 int p = lower_bound(A, A + Len, data) - A;
27 if(p == Len)
28 Len++;
29 A[p] = data;
30 }
31 }
32 printf("%d\n", Len);
33 }
34 return 0;
35 }
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