HDU 2594 Simpsons’ Hidden Talents KMP

题目的地址： http://acm.hdu.edu.cn/showproblem.php?pid=2594

题意：给你两个字符串s1,s2,让你求一个最大长度的子串t，t是s1的前缀，并且是s2的后缀，输出t和t的长度，如果不存在，直接输出0.

1、直接求next

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <list>
#include <deque>
#include <queue>
#include <iterator>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <cctype>
using namespace std;

typedef long long LL;
const int N=50005;
const LL II=100000000;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);

int next[2*N],len;
char str[2*N],xh[N];

void getnext(char *p)
{
    int j=0,k=-1;
    next[0]=-1;
    while(j<len)//len是p的长度
    {
        if(k==-1||p[j]==p[k])
        {
            j++;    k++;
            next[j]=k;
        }
        else
            k=next[k];
    }
}

int main()
{
    int i,j,T;
    while(scanf("%s%s",str,xh)!=EOF)
    {
        int len1=strlen(str),len2=strlen(xh);
        len=len1+len2;
        strcat(str,xh);
        getnext(str);
        while(next[len]>len1||next[len]>len2)
        {
            len=next[len];
        }
        str[next[len]]='';
        if(next[len]==0)
            printf("0
");
        else
            printf("%s %d
",str,next[len]);
    }
    return 0;
}

2、KMP匹配，将s1作为模式串，将s2作为主串，直接kmp

#include <cstdio>
#include <cstring>
using namespace std;

const int N = 50002;
char str1[N];
char str2[N];
int next[N];

void get_next(int len_1);
int kmp_search(int len_1, int len_2);

int main()
{
    int len;
    while(scanf("%s%s", str1, str2) != EOF)
    {
        int len_1 = strlen(str1);
        int len_2 = strlen(str2);
        get_next(len_1);
        len = kmp_search(len_1, len_2);
        if(len == 0)
        {
            printf("0
");
        }
        else
        {
            for(int i = 0; i < len; i++)
            {
                printf("%c", str1[i]);
            }
            printf(" %d
", len);
        }
    }
    return 0;
}

void get_next(int len_1)
{
    int i = 0;
    int j = -1;
    next[i] = -1;
    while(i < len_1)
    {
        if(j == -1 || str1[j] == str1[i])
        {
            i++;
            j++;
            if(str1[i] == str1[j])
            {
                next[i] = next[j];
            }
            else
            {
                next[i] = j;
            }
        }
        else
        {
            j = next[j];
        }
    }
}

int kmp_search(int len_1, int len_2)
{
    int i = 0;
    int j = 0;
    while(i < len_2)
    {
        if(j == -1 || str1[j] == str2[i])
        {
            i++;
            j++;
        }
        else
        {
            j = next[j];
        }
    }
    if(j == -1)
    {
        return 0;
    }
    if(j == 0)
    {
        if(str1[0] == str2[len_2 - 1])
        {
            return 1;
        }
        else
        {
            return 0;
        }
    }
    else
    {
        return j;
    }
}