BUPT 652 confusing problem (AC automation + digital DP)

Reprinted please indicate the source, thank youHttp://blog.csdn.net/acm_cxlove/article/details/7854526
By --- cxlove

Give an interval and ask how many numbers in the interval do not contain digits 0, and at least once A or B appears.

Http://acm.bupt.edu.cn/onlinejudge/newoj/showProblem/show_problem.php? Problem_id = 652

Since there are only two mode strings, A and B can actually be matched directly, although I won't

I still need to use the AC + digital DP. I did the zoj BCD code yesterday and planned to solve this problem with enthusiasm. As a result, I fell down for a long time and cried ~~~

I do not have sufficient understanding of the digital DP, and I still have a memory-based search.

I feel that the details of the digital DP are not easy to handle, for example, the leading 0, the upper and lower bounds, and so on. Now I especially feel the leading 0 problem.

Because this question cannot be 0, you still need to pay attention, but the high level can be 0, that is, there can be a leading 0 ???

I have been entangled in the memory problem. The result still uses 4 dimensions, DP [I] [J] [k] [l] to indicate that the length is I. Currently, it is at the J node of the automatic machine, k indicates whether all the high positions are 0, and l indicates whether a specified string is displayed.

# Include <iostream> # include <cstdio> # include <map> # include <cstring> # include <cmath> # include <vector> # include <algorithm> # include <set> # include <string> # include <queue> # define INF 1 <30 # define M 60005 # define n 10005 # define maxn 300005 # define EPS 1e-10 # define zero () FABS (a) <EPS # define min (A, B) (a) <(B )? (A) :( B) # define max (A, B) (a)> (B )? (A) :( B) # define Pb (a) push_back (a) # define MEM (a, B) memset (a, B, sizeof ()) # define ll long # define lson step <1 # define rson step <1 | 1 # define mod 1000000009 using namespace STD; struct trie {trie * Next [10]; trie * fail; int isword, kind;}; trie * que [m], s [m]; int idx; trie * newnode () {trie * TMP = & S [idx]; MEM (TMP-> next, null); TMP-> isword = 0; TMP-> fail = NULL; TMP-> kind = idx ++; return TMP;} void insert (trie * root, Char * s, int Len) {trie * P = root; For (INT I = 0; I <Len; I ++) {If (p-> next [s [I]-'0'] = NULL) P-> next [s [I]-'0'] = newnode (); P = p-> next [s [I]-'0'];} p-> isword = 1;} void bulid_fail (trie * root) {int head = 0, tail = 0; que [tail ++] = root; root-> fail = NULL; while (Head <tail) {trie * TMP = que [head ++]; for (INT I = 0; I <10; I ++) {If (TMP-> next [I]) {If (TMP = root) TMP-> next [I]-> fail = root; else {trie * P = TMP-> fail; while (P! = NULL) {If (p-> next [I]) {TMP-> next [I]-> fail = p-> next [I]; break ;} P = p-> fail;} If (P = NULL) TMP-> next [I]-> fail = root ;} if (TMP-> next [I]-> fail-> isword) TMP-> next [I]-> isword = TMP-> next [I]-> fail-> isword; que [tail ++] = TMP-> next [I];} else if (TMP = root) TMP-> next [I] = root; else TMP-> next [I] = TMP-> fail-> next [I] ;}}char STR [100]; ll l, R; ll DP [20] [55] [2] [2]; int bit [25], length, N; ll DFS (INT Len, int POs, int flag, I NT zero, bool limit) {If (LEN <= 0) return flag = 1; if (! Limit & flag & DP [Len] [POS] [zero] [1]! =-1) return DP [Len] [POS] [zero] [1]; If (! Limit &&! Flag & DP [Len] [POS] [zero] [0]! =-1) return DP [Len] [POS] [zero] [0]; ll ans = 0; int up = limit? Bit [Len]: 9; If (zero & Len> 1) ans + = DFS (len-1, = up); For (INT I = 1; I <= up; I ++) {ans + = DFS (len-1, s [POS]. next [I]-> kind, flag | (s [POS]. next [I]-> isword), 0, limit & (I = up);} If (! Limit) {DP [Len] [POS] [zero] [flag] = ans;} return ans;} ll slove (LL num) {length = 0; while (Num) {bit [++ length] = num % 10; num/= 10;} MEM (DP,-1); Return DFS (length, 0, 0, 1, true );} int main () {int t; freopen ("1.in"," r ", stdin); freopen (" 1.out", "W", stdout); scanf ("% d ", & T); While (t --) {idx = 0; trie * root = newnode (); scanf ("% LLD", & L, & R ); scanf ("% s", STR); insert (root, STR, strlen (STR); scanf ("% s", STR); insert (root, STR, strlen (STR); bustm_fail (Root); printf ("% LLD \ n", slove (R)-slove (L-1);} return 0 ;}