【数位DP】HDU3565-Bi-peak Number

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=3565

Problem Description

A peak number is defined as continuous digits {D0,D1 … Dn-1} (D0 > 0 and n >= 3),which exist Dm (0 < m < n - 1) satisfied Di-1 < Di (0 < i <= m) and Di > Di+1 (m <= i < n - 1).

A number is called bi-peak if it is a concatenation of two peak numbers.

The score of a number is the sum of all digits. Love8909 is crazy about bi-peak numbers. Please help him to calculate the MAXIMUM score of the Bi-peak Number in the closed interval [A,B].

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Input

The first line of the input is an integer T (T <= 1000),which stands for the number of test cases you need to solve.

Each case consists of two integers “A B” (without quotes) (0 <= A <= B < 2^64) in a single line.

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Output

For the kth case,output “Case k: v” in a single line where v is the maximum score. If no bi-peak number exists,output 0.

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Sample Input

  
  
   
   3
12121 12121
120010 120010
121121 121121
  
  

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Sample Output

  
  
   
   Case 1: 0
Case 2: 0
Case 3: 8
  
  
  
  

  
  

题目意思：

给范围[X,Y]，求范围内双峰数位数和最大值是多少。

双峰数定义就是满足一个数 可以分割成两个 /\ /\ 的形式。共有7种状态

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#define LL unsigned __int64
using namespace std;
const int maxn=30;
int Case=1;
int numa[maxn];     //  记录a的每一位;
int numb[maxn];     //  记录b的每一位;
int dp[maxn][10][7];    //  dp[i][j][k]表示第i位,切上一位是j,在k状态下的最大值;
//  当前位置,前一位数,当前所处的状态,是否是最大值;
int dfs(int pos,int pre,int stat,bool ismaxa,bool ismaxb)
{
    if(pos==0)
        return stat==6?0:-1;
    if(!ismaxa&&!ismaxb&&dp[pos][pre][stat]>=0)
        return dp[pos][pre][stat];
    int Min=ismaxa?numa[pos]:0;
    int Max=ismaxb?numb[pos]:9;
    int ans=-1;
    for(int i=Min;i<=Max;i++){
        int tmp=0;  //  临时变量标记当前状态;
        if(stat==0&&i){     //  刚进入的状态;
            tmp=1;
        }else if(stat==1){  //  上坡状态;
            if(i>pre) tmp=2;
            else tmp=-1;
        }else if(stat==2){  //  可上可下状态;
            if(i<pre) tmp=3;
            else if(i>pre) tmp=2;
            else tmp=-1;
        }else if(stat==3){  //  下状态;
            if(i>pre) tmp=4;
            else if(i<pre) tmp=3;
            else if(i==pre){
                if(i) tmp=4;
                else  tmp=-1;
            }
        }else if(stat==4){  //  第二个峰的上坡状态;
            if(i>pre) tmp=5;
            else tmp=-1;
        }else if(stat==5){  //  第二个峰的可上可下状态;
            if(i<pre) tmp=6;
            else if(i>pre) tmp=5;
            else tmp=-1;
        }else if(stat==6){  //  第二个峰的下状态(只有两个峰，所以只可以下);
            if(i<pre) tmp=6;
            else tmp=-1;
        }
        if(tmp!=-1){
            int sum=dfs(pos-1,i,tmp,ismaxa&&i==Min,ismaxb&&i==Max);
            if(sum!=-1) ans=max(ans,sum+i); //  和加上这位数;
        }
    }
    if(!ismaxa&&!ismaxb) dp[pos][pre][stat]=ans;
    return ans;
}
int solve(LL a,LL b)
{
    int pos=0;
    while(b){
        numa[++pos]=a%10;
        a/=10;
        numb[pos]=b%10;
        b/=10;
    }
    return dfs(pos,true,true);
}
int main()
{
    int t;
    LL a,b;
    scanf("%d",&t);
    memset(dp,-1,sizeof(dp));
    while(t--){
        scanf("%I64u%I64u",&a,&b);
        int tmp=solve(a,b);
        printf("Case %d: %d\n",Case++,tmp==-1?0:tmp);
    }
    return 0;
}