Codeforces 494B B. Obsessive String (DP)

Topic Links:

Codeforces 494B

Main topic:

Gives two strings, asking how many methods the first string extracts to make the substrings contain T-pattern strings.

Topic Analysis:

• Define state Dp[i] represents the number of methods by which the first I-character gets the string group that meets the requirements.
• d P[I]=d P[I?1]+ ∑ j = 0 l? 1 d P[J]+L
• Explain:
  
  • First Dp[i-1] represents the case that does not re-construct a new substring, then the front if selected, then the current bit must be selected, the previous vacancy, then the current position must be vacant, so the number of cases is equal to dp[i-1]
  • After considering L is to ensure that the current paragraph contains the maximum number of T, fixed the right position, and then enumerate the left position, because it will be more than the previous situation of the case of only the string itself, so it is dp[j]+1, and then extracted to become the final plus L.
    

AC Code:

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define MAX 100007using namespace STD;CharS[max],t[max];intMark[max];Const intMoD =1e9+7;intDp[max];intSum[max];voidGet_next (CharP[],intNext []) {inti =0, k =-1, Len =strlen(p); next[0] = -1; while(I < Len)if(k = =-1|| P[i] = = P[k]) i++,k++,next[i]=k;Elsek = Next[k];}voidMatch (CharS[],CharP[]) {memset(Mark,0,sizeof(Mark));intNext[max]; Get_next (P, next);intLen1 =strlen(s);intLen2 =strlen(p);inti =0, j =0; while(I < LEN1) {if(j = =-1|| S[i] = = P[j]) i++, j + +;Elsej = Next[j];if(j = = len2) Mark[i] = i-len2+1; }}intMain () { while( ~scanf('%s ', s)) {scanf('%s ', t);        Match (S, T); sum[0] = dp[0] =0;intn =strlen(s); for(inti =1; I <= N; i++)if(!mark[i]) mark[i] = mark[i-1]; for(inti =1; I <= N; i++) {Dp[i] = dp[i-1];intL = mark[i];if(!l)Continue; Dp[i] + = (sum[l-1]+L)%mod;            Dp[i]%= mod; Sum[i] = sum[i-1] + dp[i];        Sum[i]%= mod; }printf("%d\n", Dp[n]); }}

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Codeforces 494B B. Obsessive String (DP)