UVA 10534 Wavio Sequence DP LIS

题意:求一个波浪子序列,就是是前一半是上升子序列,后一半是下降子序列(子序列的长度必须为奇数)。

分别从左右两个方向求LIS,然后在统计最大值就行了

 //#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define pb(a) push_back(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
void debug()
{
#ifdef ONLINE_JUDGE
#else
    freopen("d:\in.txt","r",stdin);
   // freopen("d:\out1.txt","w",stdout);
#endif
}
char getch()
{
    char ch;
    while((ch=getchar())!=EOF)
    {
        if(ch!=' '&&ch!='
')return ch;
    }
    return EOF;
}
int dp1[10010],dp2[10010];
int da[10010];
int d[10010];
int work1(int n)
{
    int maxx=0;
    for(int i=1;i<=n;i++)
    {
        int pos=lower_bound(d+1,d+1+maxx,da[i])-d;
        dp1[i]=pos;
        d[pos]=da[i];
        maxx=max(maxx,pos);
    }
    return 0;
}
int work2(int n)
{
    int maxx=0;
    for(int i=n;i>=1;i--)
    {
        int pos=lower_bound(d+1,d+1+maxx,da[i])-d;
        dp2[i]=pos;
        d[pos]=da[i];
        maxx=max(maxx,pos);
    }
    return 0;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&da[i]);
        int maxx=0;
        work1(n);
        work2(n);
        for(int i=1;i<=n;i++)
        {
            int x=min(dp1[i],dp2[i]);
            maxx=max(maxx,x*2-1);
        }
        printf("%d
",maxx);
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/BMan/p/3250319.html