LIS UVA 10534 Wavio Sequence

题目传送门

 1 /*
 2     LIS：应用，nlogn的做法，首先从前扫到尾，记录每个位置的最长上升子序列，从后扫到头同理。
 3             因为是对称的，所以取较小值*2-1再取最大值
 4 */
 5 /************************************************
 6 * Author        :Running_Time
 7 * Created Time  :2015-8-5 21:38:32
 8 * File Name     :UVA_10534.cpp
 9  ************************************************/
10
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29
30 #define lson l, mid, rt << 1
31 #define rson mid + 1, r, rt << 1 | 1
32 typedef long long ll;
33 const int MAXN = 1e4 + 10;
34 const int INF = 0x3f3f3f3f;
35 const int MOD = 1e9 + 7;
36 int a[MAXN];
37 int d[MAXN];
38 int dp[MAXN], dp2[MAXN];
39
40 int main(void)    {     //UVA 10534 Wavio Sequence
41     int n;
42     while (scanf ("%d", &n) == 1)   {
43         for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
44         memset (d, 0, sizeof (d));
45         memset (dp, 0, sizeof (dp));
46         memset (dp2, 0, sizeof (dp2));
47         d[1] = a[1];    int len = 1;    dp[1] = 1;
48         for (int i=2; i<=n; ++i)    {
49             if (d[len] < a[i])  d[++len] = a[i];
50             else    {
51                 int j = lower_bound (d+1, d+1+len, a[i]) - d;
52                 d[j] = a[i];
53             }
54             dp[i] = len;
55         }
56         d[1] = a[n];    int len2 = 1;   dp2[n] = 1;
57         for (int i=n-1; i>=1; --i)    {
58             if (d[len2] < a[i]) d[++len2] = a[i];
59             else    {
60                 int j = lower_bound (d+1, d+1+len2, a[i]) - d;
61                 d[j] = a[i];
62             }
63             dp2[i] = len2;
64         }
65         int ans = 0;
66         for (int i=1; i<=n; ++i)    {
67             ans = max (ans, min (dp[i], dp2[i]) * 2 - 1);
68         }
69         printf ("%d\n", ans);
70     }
71
72     return 0;
73 }