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  • POJ 2452 Sticks Problem

    RMQ+二分。。。。
    枚举 i  ,找比 i 小的第一个元素,再找之间的第一个最大元素。。。。。

                      Sticks Problem
    Time Limit: 6000MS   Memory Limit: 65536K
    Total Submissions: 9338   Accepted: 2443

    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

     
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 
     5 using namespace std;
     6 
     7 int dp_max[60000][16],dp_min[60000][16],a[60000],n;
     8 
     9 int _max(int l,int r)
    10 {
    11     if(a[l]>=a[r]) return l;
    12     else return r;
    13 }
    14 
    15 int _min(int l,int r)
    16 {
    17     if(a[l]<=a[r]) return l;
    18     else return r;
    19 }
    20 
    21 void Init()
    22 {
    23     for(int i=1;i<=n;i++)
    24         dp_max[i][0]=dp_min[i][0]=i;
    25     for(int j=1;(1<<j)<=n;j++)
    26     {
    27         for(int i=1;i+(1<<j)-1<=n;i++)
    28         {
    29             int m=i+(1<<(j-1));
    30             dp_max[i][j]=_max(dp_max[i][j-1],dp_max[m][j-1]);
    31             dp_min[i][j]=_min(dp_min[i][j-1],dp_min[m][j-1]);
    32         }
    33     }
    34 }
    35 
    36 int rmq_min(int L,int R)
    37 {
    38     int k=0;
    39     while(L+(1<<k)-1<=R) k++; k--;
    40     return _min(dp_min[L][k],dp_min[R-(1<<k)+1][k]);
    41 }
    42 
    43 int rmq_max(int L,int R)
    44 {
    45     int k=0;
    46     while(L+(1<<k)-1<=R) k++; k--;
    47     return _max(dp_max[L][k],dp_max[R-(1<<k)+1][k]);
    48 }
    49 
    50 int main()
    51 {
    52     while(scanf("%d",&n)!=EOF)
    53     {
    54         for(int i=1;i<=n;i++)
    55             scanf("%d",a+i);
    56         Init();
    57         int ans=0,r,k;
    58         for(int i=1;i+ans<n;i++)
    59         {
    60             int low=i+1,high=n,mid;
    61             while(low<=high)
    62             {
    63                 if(low==high) break;
    64                 mid=(low+high)>>1;
    65                 if(a[i]<a[rmq_min(low,mid)])
    66                     low=mid+1;
    67                 else
    68                     high=mid;
    69             }
    70             r=low;
    71             k=rmq_max(i,r);
    72             if(a[i]<a[k])
    73                 ans=max(ans,k-i);
    74         }
    75         if(ans==0) printf("-1
    ");
    76         else printf("%d
    ",ans);
    77     }
    78     return 0;
    79 }
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  • 原文地址:https://www.cnblogs.com/CKboss/p/3358080.html
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