HDU2594 Simpsons’ Hidden Talents 字首與字尾轉化成用s1去匹配s2
http://acm.hdu.edu.cn/showproblem.php?pid=2594
Simpsons’ Hidden TalentsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15076 Accepted Submission(s): 5168 Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Input Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.
Output Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. If the longest such string is the empty string, then the output should be 0.
Sample Input clinton homer riemann marjorie
Sample Output 0 rie 3
Source HDU 2010-05 Programming Contest
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//輸入s1,s2 //用s2當做目標串，s1去匹配，看最大匹配長度j； #include <iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> using namespace std; const int MAXN=50000+1000; char T[MAXN],P[MAXN]; int f[MAXN],ex[MAXN]; int n,m; void getFail() { f[0]=f[1]=0; for(int i=1;i<m;i++) { int j=f[i]; while(j && P[i]!=P[j]) j=f[j]; f[i+1] = (P[i]==P[j])?j+1:0; } } void find() { int j=0; for(int i=0;i<n;i++) { while(j && T[i]!=P[j]) j=f[j]; if(T[i]==P[j])j++; ex[i]=j;//最大匹配長度 } } int main() { while(scanf("%s %s",P,T)==2) { n=strlen(T); m=strlen(P); getFail(); find(); if(ex[n-1]==0) printf("0\n"); else { for(int i=0;i<ex[n-1];i++) printf("%c",P[i]); printf(" %d\n",ex[n-1]); } } return 0; }