[Digital DP] HDU 3565 bi-peak number

Question:

For the range [x, y], determine the number of dual-peaks and the maximum value in the range.

A double-peak number is defined as a number that can be divided into two //// \ formats.

Ideas:

The number before the DP [site] [cur] [OK] site bit is that the cur status is OK.

There are 7 OK types

0: all preceding values are 0.

1: the first peak and only one number

2: the first peak is on the top (up or down)

3: the first peak is at the bottom of the peak (you can enter the next peak or continue)

4: The same 1 is the second peak

5: Same 2 is the second peak

6: Same as 3, but cannot enter the next peak

Code:

# Include "cstdlib" # include "cstdio" # include "cstring" # include "cmath" # include "queue" # include "algorithm" # include "iostream" using namespace STD; # define ll unsigned _ int64int DP [30] [10] [7]; int numx [30], numy [30]; int DFS (INT site, int cur, int OK, int FA, int FB) // judge {If (Site = 0) Return OK = 6? 0:-1; // status 6 indicates the number of valid if (! Fa &&! FB &&~ DP [site] [cur] [OK]) return DP [site] [cur] [OK]; // none of them are boundary int min = fa? Numx [site]: 0; // upper bound int max = FB? Numy [site]: 9; // lower bound int ans =-1; // Initial Value for (INT I = min; I <= max; I ++) {int TEP = 0; If (OK = 0 & I) TEP = 1; // deprecated 0 else if (OK = 1) {if (I> cur) TEP = 2; // go up to else TEP =-1; // cannot go} else if (OK = 2) {if (I> cur) TEP = 2; // continue with else if (I = cur) TEP =-1; // equal cannot take else TEP = 3; // down} else if (OK = 3) {if (I> cur) TEP = 4; // jump to the second else if (I = cur) // 0 cannot jump if it is equal, because 0 {if (I) TEP = 4; else TEP =-1;} else TEP = 3; // continue} else if (OK = 4 )// Same as {if (I> cur) TEP = 5; else TEP =-1;} else if (OK = 5) {if (I> cur) TEP = 5; else if (I = cur) TEP =-1; else TEP = 6;} else if (OK = 6) {if (I >= cur) TEP =-1; // you can only skip else TEP = 6;} If (TEP! =-1) {int sum = DFS (site-1, I, TEP, fa & I = min, FB & I = max ); // The maximum value after this end, if (sum! =-1) ans = max (ANS, sum + I); // Add this ratio} If (! Fa &&! FB) DP [site] [cur] [OK] = ans; // return ans;} int main () {int T, CAS = 1; cin> T; memset (DP,-1, sizeof (DP); While (t --) {ll X, Y; scanf ("% i64u % i64u ", & X, & Y); // note that to power 2 ^ 64, use unsigned 64-bit to read int CNT = 0; while (y) {CNT ++; numx [CNT] = x % 10; X/= 10; numy [CNT] = Y % 10; y/= 10;} int ans = DFS (CNT, 0, 0, 1, 1); printf ("case % d: % d \ n", CAS ++, ANS =-1? 0: ANS);} return 0 ;}

[Digital DP] HDU 3565 bi-peak number