HDU 3565 Bi-peak Number [数位DP]

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].

题意：定义一种双峰数，这个数由两座山峰构成，每一座山峰至少三个数，且第一位不为0，保证其先递增，后递减，每一部分的数不少于2个（最高数共用），每一个双峰数能够得到的分数是每一位上的数字相加，问区间[A,B]内所有的双峰数的分数中的最大值。

解法：很明显是数位DP，但是状态的定义需要仔细考虑一下，dp[i][j][k] 表示处理到第I位（从高往低），前一位的数字为J，且当前状态为K时，之后的所有可能存在的数，剩下所有位（I位之后）之和的最大值。K我定义了6个状态，分别是第一次上坡，顶峰，下坡以及第二次的，然后记忆化搜索即可，记忆化的时候还需要增加三维来表示之前取得数的和，以及当前是否达到询问的上界或下界。

代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#include<iostream>#include<stdlib.h>#include<set>#include<map>#include<queue>#include<vector>#include<bitset>#pragma comment(linker, "/STACK:1024000000,1024000000")template <class T>bool scanff(T &ret){ //Faster Input    char c; int sgn; T bit=0.1;    if(c=getchar(),c==EOF) return 0;    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();    sgn=(c=='-')?-1:1;    ret=(c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');    if(c==' '||c=='\n'){ ret*=sgn; return 1; }    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;    ret*=sgn;    return 1;}#define inf 1073741823#define llinf 4611686018427387903LL#define PI acos(-1.0)#define lth (th<<1)#define rth (th<<1|1)#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)#define mem(x,val) memset(x,val,sizeof(x))#define mkp(a,b) make_pair(a,b)#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define pb(x) push_back(x)using namespace std;typedef unsigned long long  ll;typedef pair<int,int> pii;int numd[22],numu[22];int dp[22][11][10];int dfs(int pos,int pre,int s,int sum,int limd,int limu){    if(pos==0){        if(s==6)return sum;        else return 0;    }    if(!limd&&!limu&&~dp[pos][pre][s])return dp[pos][pre][s]+sum;    int d=0,u=9;    if(limd)d=numd[pos];    if(limu)u=numu[pos];    int ans=0;    rep(i,d,u){        if(i==0&&s==0)ans=max(ans,dfs(pos-1,0,0,0,limd&&i==d,limu&&i==u));        else{            if(i!=0){                if(s==0)ans=max(ans,dfs(pos-1,i,1,sum+i,limd&&i==d,limu&&i==u));                if(s==3)ans=max(ans,dfs(pos-1,i,4,sum+i,limd&&i==d,limu&&i==u));            }            if(i>pre){                if(s==1||s==4)ans=max(ans,dfs(pos-1,i,s+1,sum+i,limd&&i==d,limu&&i==u));                if(s==2||s==5)ans=max(ans,dfs(pos-1,i,s,  sum+i,limd&&i==d,limu&&i==u));            }            if(i<pre){                if(s==2||s==5)ans=max(ans,dfs(pos-1,i,s+1,sum+i,limd&&i==d,limu&&i==u));                if(s==3||s==6)ans=max(ans,dfs(pos-1,i,s,  sum+i,limd&&i==d,limu&&i==u));            }        }    }    if(!limd&&!limu)dp[pos][pre][s]=ans-sum;    return ans;}void solve(ll x,ll y){    mem(numd,0);mem(numu,0);    for(;x;x/=10)numd[++numd[0]]=x%10;    for(;y;y/=10)numu[++numu[0]]=y%10;    printf("%d\n",dfs(numu[0],0,0,0,1,1));}int main(){    mem(dp,-1);    tdata{        ll x,y;        scanff(x);scanff(y);        printf("Case %d: ",cas);        solve(x,y);    }    return 0;}/*61546 6898591659566 165989811323 1659491313 1549749121 132656416132661 13466421*/