【数位DP】HDU3565-Bi-peak Number

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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].

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.

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.

Sample Input

312121 12121120010 120010121121 121121

Sample Output

Case 1: 0Case 2: 0Case 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 __int64using 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,0,0,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;}

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