UVA 10534 Wavio Sequence

题解:

根据最长上升子序列的nlog(n)做法，求出以每个点结尾的最长上升子序列。

然后反过来再求一次

然后取最小值 * 2 - 1 即可

代码:

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pii pair<int,int>
const int INF = 1000000000;
const int maxn = 10050;
const int M = 4000;

int n;
int a[ maxn ], b[ maxn ];
int num[ maxn ],num2[ maxn ];

int bs( int s, int e, int x )
{
    int l = s, r = e, goal = e + 1;
    while( l <= r )
    {
        int m =( l + r ) >> 1;
        if( b[ m ] >= x )
        {
            goal = m;
            r = m -1;
        }
        else l = m + 1;
    }
    return goal;
}

int main()
{
    int n;
    while( ~scanf( "%d", &n ) )
    {
        for( int i = 1; i <= n; i ++ ) scanf( "%d", &a[ i ] );
        int tmp = 1;
        b[ tmp ] = a[ tmp ];
        num[ tmp ] = 1;
        for( int i = 2; i <= n; i ++ )
        {
            int pos =bs( 1, tmp, a[ i ] );
            b[ pos ] = a[ i ];
            num[ i ] = pos;
            if( pos == tmp + 1 ) tmp ++;
        }
        reverse( a + 1, a + n + 1 );
        tmp = 1;
        b[ tmp ] = a[ tmp ];
        num2[ tmp ] = 1;
        for( int i = 2; i <= n; i ++ )
        {
            int pos =bs( 1, tmp, a[ i ] );
            b[ pos ] = a[ i ];
            num2[ i ] = pos;
            if( pos == tmp + 1 ) tmp ++;
        }
        int G = 0;
        for( int i = 1; i <= n; i ++ ) G = max( G, min( num[ i ], num2[ n + 1 - i ] ) * 2 - 1 );
        printf( "%d\n", G );
    }
    return 0;
}