uva10534---Wavio Sequence

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=17&problem=1475&mosmsg=Submission+received+with+ID+13386999

Wavio is a sequence of integers. It has some interestingproperties.

·  Wavio is of odd length i.e. L = 2*n+ 1.

·  The first (n+1) integers ofWavio sequence makes a strictly increasing sequence.

·  The last (n+1) integers of Waviosequence makes a strictly decreasing sequence.

·  No two adjacent integers are same in aWavio sequence.

For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Waviosequence of length9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not avalid wavio sequence. In this problem, you will be given a sequence ofintegers. You have to find out the length of the longest Wavio sequence whichis a subsequence of the given sequence. Consider, the given sequence as :

1 2 3 2 1 2 3 4 32 1 5 4 1 2 3 2 2 1.

Here the longest Wavio sequence is : 1 2 3 4 5 4 3 2 1. So, the outputwill be9.

 

Input

The input filecontains less than 75 test cases. The description of each test case isgiven below: Input is terminated by end of file.

 

Each set starts with a postiveinteger, N(1<=N<=10000). In next few lines there will beNintegers.

 

Output

For each set of input print the length of longest waviosequence in a line.

Sample Input                                  Output for Sample Input

10

1 2 3 4 5 4 3 2 1 10

19

1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1

5

1 2 3 4 5

9

9

1

 

这道题主要是LIS的应用，把串正反走一遍就行，代码中LIS（）函数是精华。

代码实现：

#include <iostream>#include <cstring>using namespace std;#define N 10001int dp1[10001],dp2[10001],a[10001],b[10001];int t;void LIS(int dp[],int a[]){    int stack[N];    int top=0;    stack[top]=-99999999;    for(int i=1; i<=t; i++)    {        //如果a[i]>栈顶部元素，则压栈        if(a[i]>stack[top])        {            stack[++top]=a[i];            dp[i]=top;        }        //如果a[i]不大于栈顶部元素，则二分查找第一个比a[i]大的元素        else        {            int l=1,r=top;            while(l<=r)            {                int mid=(l+r)>>1;                if(a[i]>stack[mid])                {                    l=mid+1;                }                else                    r=mid-1;            }            //替换a[i]            stack[l]=a[i];            dp[i]=l;        }    }}int main(){    while(cin>>t)    {        for(int i=1;i<=t;i++)        {           cin>>a[i];           b[t-i+1]=a[i];        }        memset(dp1,0,sizeof(dp1));        memset(dp2,0,sizeof(dp2));        LIS(dp1,a);        LIS(dp2,b);        int maxx=-1,ans=0;        for(int i=1;i<=t;i++)        {            ans=min(dp1[i],dp2[t-i+1])*2-1;            if(ans>maxx)            {                maxx=ans;            }        }        cout<<maxx<<endl;    }    return 0;}