POJ 2452 Sticks Problem

最新推荐文章于 2020-04-17 19:26:40 发布

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本文介绍了一种使用RMQ（Range Minimum Query）和二分查找的算法来解决棍子问题，该问题涉及到找出两根棍子之间长度最大的棍子。通过输入棍子的数量和每根棍子的长度，输出满足条件的最大距离。该方法利用了RMQ的数据结构和二分搜索的高效性，解决了问题并给出了详细的实现步骤。

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

Sample Output

1
-1

Source

POJ Monthly,static

 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\n");
76         else printf("%d\n",ans);
77     }
78     return 0;
79 }