UVA 10534 Wavio Sequence(LIS)

题意：给定一个长度为n的整数序列，求一个最长子序列，使得该序列的长度为2*k+1，前k+1个数严格递增，后k+1个数严格递减。

分析：用O(nlogn)分别求出递增序列和递减序列，再O(n)求在某个位置符合条件的序列长度。

View Code 
 1 /*
 2 Author:Zhaofa Fang
 3 Lang:C++
 4 */
 5 #include <cstdio>
 6 #include <cstdlib>
 7 #include <sstream>
 8 #include <iostream>
 9 #include <cmath>
10 #include <cstring>
11 #include <algorithm>
12 #include <string>
13 #include <utility>
14 #include <vector>
15 #include <queue>
16 #include <stack>
17 #include <map>
18 #include <set>
19 using namespace std;
20 
21 typedef long long ll;
22 #define DEBUG(x) cout<< #x << ':' << x << endl
23 #define REP(i,n) for(int i=0;i < (n);i++)
24 #define FOR(i,s,t) for(int i = (s);i <= (t);i++)
25 #define PII pair<int,int>
26 #define PB push_back
27 #define MP make_pair
28 #define FI first
29 #define SE second
30 #define lowbit(x) (x&(-x))
31 #define INF (1<<30)
32 
33 const int maxn = 10011;
34 int a[maxn];
35 int g1[maxn],g2[maxn];
36 int d1[maxn],d2[maxn];
37 int mx1[maxn],mx2[maxn];
38 int main()
39 {
40     //freopen("in","r",stdin);
41     int n;
42     while(~scanf("%d",&n))
43     {
44         for(int i=0;i<n;i++)scanf("%d",&a[i]);
45         for(int i=1;i<=n;i++)g1[i] = g2[i] = INF;
46         memset(mx1,0,sizeof(mx1));
47         memset(mx2,0,sizeof(mx2));
48         for(int i=0;i<n;i++)
49         {
50             int k = lower_bound(g1+1,g1+n+1,a[i]) - g1;
51             g1[k] = a[i];
52             d1[i] = k;
53         }
54         for(int i=n-1;i>=0;i--)
55         {
56             int k = lower_bound(g2+1,g2+n+1,a[i]) - g2;
57             g2[k] = a[i];
58             d2[i] = k;
59         }
60         mx1[0] = d1[0];
61         for(int i=1;i<n;i++)mx1[i] = max(mx1[i-1],d1[i]);
62         for(int i=n-1;i>=0;i--)mx2[i] = max(mx2[i+1],d2[i]);
63         int ans = 0;
64         for(int i=0;i<n;i++)
65         {
66             int tmp = min(mx1[i],mx2[i]);
67             ans = max(ans,(tmp-1)*2+1);
68             //DEBUG(i);
69             //DEBUG(mx1[i]);DEBUG(mx2[i]);DEBUG(ans);
70         }
71         printf("%d\n",ans);
72     }
73     return 0;
74 }
posted @ 2012-11-10 15:11  發_  阅读(196)  评论(0)  编辑  收藏  举报