【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].

 

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

3
12121 12121
120010 120010
121121 121121

 

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,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;
}