hdu 4105 Electric wave DP

Problem Description

  
  

   
   Ali was doing a physic experiment which requires him to observe an electric wave. He needs the height of each peak value and valley value for further study (a peak value means the value is strictly larger than its neighbors and a valley value means the value is strictly smaller than its neighbors). He did write these numbers down but he was too careless that he wrote them in a line without separations, such as “712495” may represent “7 12 4 9 5”. The only information he can remember was:
1. The data begins with a valley value
2. Each value is either a peak value or a valley value
Now he wants to insert blanks to make the data valid. If multiple solutions exist, he will choose the one with more blanks.
  
  

 

Input

  
  

   
   The input consists several testcases.
The first line contains one integer N (1 <= N <= 100), the length of the data.
The second line contains one string S, the data he recorded.
S contains only digits.
  
  

 

Output

  
  

   
   Print one integer, the maximum number of blanks he can insert.
  
  

 

Sample Input

  
  

   
   6
712495
  
  

 

Sample Output

  
  

   
   4

   
   

    
    

     
     Hint
    
    
The separated data may have leading zeros.
   
   
  
  

//

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<string>
using namespace std;
int n;
int a[110];
int f[110][2][110];
int compare(int l1,int r1,int l2,int r2)
{//cout<<l1<<".."<<r1<<".."<<l2<<".."<<r2<<endl;
    while(a[l1]==0&&l1<=r1) l1++;
    while(a[l2]==0&&l2<=r2) l2++;
    int len1=r1-l1+1;
    int len2=r2-l2+1;
    if(len1>len2) return 1;
    else if(len1<len2) return -1;
    for(int k=0;k<len1;k++)
    {
        if(a[l1+k]>a[l2+k]) return 1;
        else if(a[l1+k]<a[l2+k]) return -1;
    }
    return 0;
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        char str[110];scanf("%s",str);
        for(int i=1;i<=n;i++) a[i]=str[i-1]-'0';
        memset(f,-1,sizeof(f));
        f[1][0][1]=0;
        for(int i=2;i<=n;i++)
        {
            f[i][0][i]=0;
            for(int j=1;j<i;j++)
            {
                for(int l=1;l<=i-j;l++)
                {//cout<<compare(i-j+1,i,i-j-l+1,i-j)<<"........"<<endl;
                    if(compare(i-j+1,i,i-j-l+1,i-j)<0)
                    {
                        if(f[i-j][1][l]==-1) continue;
                        f[i][0][j]=max(f[i][0][j],f[i-j][1][l]+1);
                    }
                    if(compare(i-j+1,i,i-j-l+1,i-j)>0)
                    {
                        if(f[i-j][0][l]==-1) continue;
                        f[i][1][j]=max(f[i][1][j],f[i-j][0][l]+1);
                    }
                }
            }
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        {//cout<<f[n][0][i]<<"..."<<f[n][1][i]<<endl;
            ans=max(ans,f[n][0][i]);
            ans=max(ans,f[n][1][i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}
/*


f[i][2][j]表示前i个数以j个数结尾(0:低谷,1:高地)的插入最多空格数目




f[i][0][j]=max(f[i-k][1][l])+1;
f[i][1][j]=max(f[j-k][0][l])+1;




3
101
*/