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  • 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 }
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  • 原文地址:https://www.cnblogs.com/fzf123/p/2764036.html
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