POJ 2452 Sticks Problem

Sticks Problem

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 8623		Accepted: 2222

Description

Xuanxuan has n sticks of different length. One day, she puts all her sticks in a line, represented by S1, S2, S3, ...Sn. After measuring the length of each stick Sk (1 <= k <= n), she finds that for some sticks Si and Sj (1<= i < j <= n), each stick placed between Si and Sj is longer than Si but shorter than Sj.

Now given the length of S1, S2, S3, …Sn, you are required to find the maximum value j - i.

Input

The input contains multiple test cases. Each case contains two lines.
Line 1: a single integer n (n <= 50000), indicating the number of sticks.
Line 2: n different positive integers (not larger than 100000), indicating the length of each stick in order.

Output

Output the maximum value j - i in a single line. If there is no such i and j, just output -1.

Sample Input

4

5 4 3 6

4

6 5 4 3

Sample Output

1

-1

Source

POJ Monthly,static

//这题对于我来说，难点是这个二分查找、第一次知道原来可以这样二分查找

//求最大的 j-i(j>i) 对于任意的 i<k<j 有Si<k<Sj成立
//先求出离Sj最近，但比Sj大的St位置，然后i就是St~Sj区间最小值了；

#include <iostream>

#include <stdio.h>

#include <algorithm>

#include <string.h>

#include <queue>

#define lson l,m,k<<1

#define rson m+1,r,k<<1|1

using namespace std;

#define N 50010

struct node

{

    int min,id;

    int max;

}st[N<<2];

int rc[N];

int rid[N<<1];

void up(int &k)

{

   st[k].max=max(st[k<<1].max,st[k<<1|1].max);

   st[k].min=min(st[k<<1].min,st[k<<1|1].min);

}

void build(int l,int r,int k)

{

    if(l==r)

    {

        scanf("%d",&rc[l]);

        st[k].max=st[k].min=rc[l];

        rid[rc[l]]=l;

        return ;

    }

    int m=(l+r)>>1;

    build(lson);

    build(rson);

    up(k);

}

int Find_Max(int L,int R,int l,int r,int k)

{

    if(L<=l&&R>=r)

     return st[k].max;

    int m=(l+r)>>1,t1=0,t2=0;

    if(L<=m) t1=Find_Max(L,R,lson);

    if(R>m)  t2=Find_Max(L,R,rson);

    return max(t1,t2);

}

int Find_Min(int L,int R,int l,int r,int k)

{

    if(L<=l&&R>=r)

     return st[k].min;

    int m=(l+r)>>1,t1=1000000,t2=1000000;

    if(L<=m) t1=Find_Min(L,R,lson);

    if(R>m)  t2=Find_Min(L,R,rson);

    return min(t1,t2);

}

int n;

int bf(int R)

{

    int l=1,r=R,m;

    int temp;R++;

    while(l<r)

    {

        m=(l+r)>>1;

        temp=Find_Max(m+1,r,1,n,1);

        if(temp>rc[R])

          l=m+1;

        else

          r=m;

    }

    return l;

}

int main()

{



    int j,k,t,var;

    int Max,id;

    while(scanf("%d",&n)!=EOF)

    {

        Max=0;

        build(1,n,1);

        j=n;

        while(j>1)

        {

           var=rc[j];

           id=bf(j-1);

           if(rc[id]>rc[j])

             k=id+1;

           else

             k=id;

           var=Find_Min(k,j,1,n,1);

           id=rid[var];

           Max=max(j-id,Max);

           j=id-1;

        }

        if(Max)

          printf("%d\n",Max);

        else

          printf("-1\n");

    }

    return 0;

}