hdu 2594 Simpsons’ Hidden Talents KMP应用

Simpsons’ Hidden Talents

Problem Description

Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.

Sample Input

clinton homer

riemann marjorie

 

Sample Output

0

rie 3

 

思路:要求的是s1的最长前缀是s2的后缀；那么kmp中的getfail()就是用前缀来的匹配以当前点为尾点的子串的。那么如果我们 s1 += s2;那么所谓的s2的子串，就是len的匹配在s1长度总的f[i];

理解f[i]:当你那i+1的f[i+1]时，这是匹配的就是T[i] = p[i]了；

坑点:当 |s1| < |s2|时，f[i] <= |s2| ; 还有就是TLE有时不是太慢了。。而是空间炸了。。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 50007;
char T[N<<1],p[N];
int f[N<<1];
void getfail(char* p)
{
    f[0] = f[1] = 0;
    int n = strlen(p);
    for(int i = 1;i < n;i++){
        int j = f[i];
        if(j && p[i] != p[j]) j = f[j];
        f[i+1] = (p[i] == p[j] ?j+1:0);// ｉ＋１会递推到第ｎ位
    }
}
int main()
{
    while(scanf("%s%s",T,p) == 2){
        int n = strlen(T),m = strlen(p),ans = 0;
        strcat(T,p);
        getfail(T);
        for(int j = n+m;j >= n || j >= m;j = f[j]){// n+m的f[]就是匹配后缀
            //cout<<f[j]<<" ";
            if(f[j] <= n && f[j] <= m){
                ans = f[j];
                break;
            }
        }
        if(ans) for(int i = 0;i < ans;i++) putchar(T[i]);
        if(ans) putchar(' ');
        printf("%d\n",ans);
    }
}