HDU 5414 CRB and String (greedy)

Title: http://acm.hdu.edu.cn/showproblem.php?pid=5414

Test instructions: Given the string s and T, you can select a character C in S, and then choose a character D (d! =C) Insert D into the back of C and ask if you can change s to T.

Analysis: First consider the character (string) can be inserted in the case, if the string "SD" character s after the string P (|p|>=1), as long as p[0]!= ' s ' line, such as p= "Oola", the first O is inserted in the ' s ' behind, the second also inserted in the ' s ' behind, is OK. Then the characters in S and T are matched and the subscript of the character in T is as far back as possible (from backward to forward), indicating which characters in T are fixed within S. The first character of the string in the middle of the two-mark character is then directly determined to satisfy the condition. Special cases "AAA" and "AASAA", if the next two a matches the words obviously not, in the past, when judging by the way to modify the matching value on the line.

Code:

#include <iostream> #include <cstring> #include <cstdio>using namespace std;const int maxn = 1e5+6;char S1[maxn],s2[maxn];int Match[maxn];int Main () {int ncase,len1,len2,i,j,no;scanf ("%d", &ncase), while (ncase--) { scanf ("%s%s", s1,s2); Len1=strlen (S1); Len2=strlen (S2); if (S1[0]!=s2[0] | | len1>len2) {printf ("no\n"); continue;} NO = 0;j=len2-1;for (i=len1-1;i>=1;i--) {while (j>=0 && s2[j]!=s1[i]) j--;if (j<1) {no=1;break;} match[i]=j--;} Match[0]=0;if (NO) {printf ("no\n"); continue;} for (i=0;i<len1;i++) {if (i+1<len1 && match[i+1]==match[i]+1) continue; if (S2[match[i]+1]==s1[i]) {if (i+ 1<len1 && s1[i+1]==s2[match[i]+1]) {match[i+1]=match[i]+1;continue;} No=1;break;}} if (NO) printf ("no\n"); elseprintf ("yes\n");} return 0;}

HDU 5414 CRB and String (greedy)