hdu 2594 Simpsons’ Hidden Talents

方法一：

題意：求兩個串的前綴和後綴最大的長度，
思路：將第一個串作爲模式串 第二個作爲原串，進行KMP
當 i == len1 作爲結束條件

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 50000+10;
char s1[maxn], s2[maxn];
int len1, len2;
int nextval[maxn];

void get_nextval()
{
    int i = 0, j = -1;
    nextval[0] = -1;
    while(i < len2)
    {
        if(j == -1 || s2[i] == s2[j])
        {
            if(s2[++i] != s2[++j]) nextval[i] = j;
            else nextval[i] = nextval[j];
        }
        else j = nextval[j];
    }
}

void kmp_search()
{
    get_nextval();
    int i = 0, j = 0;
    while(i < len1 && j <= len2)//i == len1 結束
    {
        if(j == -1 || s1[i] == s2[j]) ++i, ++j;
        else j = nextval[j];
        if(i == len1)
        {
            i = i-j;
            break;
        }
    }
    if(j == 0) printf("0\n");//如果沒有匹配的就 j == 0
    else
    {
        for(int k = i; k < len1; k++) printf("%c", s1[k]);
        printf(" %d\n", len1-i);
    }
}

int main()
{
    while(scanf("%s%s", s2, s1) != EOF)
    {
        len1 = strlen(s1);
        len2 = strlen(s2);
        kmp_search();
    }
}

方法二：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 100000+10;
char a[maxn], b[maxn];
int len1, len2,len3, nextval[maxn];

void get_nextval()
{
    int i = 0, j = -1;
    nextval[i] = -1;
    while(i < len1+len2)
    {
        if(j == -1 || a[i] == a[j])
        {
            if(a[++i] != a[++j]) nextval[i] = j;
            else nextval[i] = nextval[j];
        }
        else j = nextval[j];
    }
}

int main()
{
    while(scanf("%s%s", a, b) != EOF)
    {
        len1 = strlen(a);
        len2 = strlen(b);
        len3 = len1 + len2;
        strcat(a, b);
        get_nextval();
        int x = nextval[len3];
        if(x)
        {
            if(x > len1 || x > len2)
            {
                int m;
                if(len1 > len2) m = len2;
                else m = len1;
                for(int i = 0; i < m; i++)
                    printf("%c", a[i]);
                printf(" %d\n", m);
            }
            else
            {
                for(int i = 0; i < x; i++)
                    printf("%c", a[i]);
                printf(" %d\n", x);
            }
        }
        else printf("0\n");
    }
}