hdu2594_VJkuangbin16kmp_Simpsons’ Hidden Talents

题目大意:给定两个字符串p, s,问最长的p的前缀是s的后缀,输出该前缀和其长度

用p去匹配s,kmp的while退出条件是s的迭代器i >= n,则退出匹配循环后p的迭代器j就为所求前缀的尾部;

#include <cstdio>
#include <cstring>

using namespace std;
const int maxn = 50010;
int next[maxn];
char s[maxn], p[maxn];
int n, m;

void getnext()
{
    int i = 0, j = -1;
    next[0] = -1;
    while(i < m)
    {
        if(j==-1 || p[i]==p[j])
        {
            ++i;
            ++j;
            next[i] = j;
        }
        else j = next[j];
    }
}

int kmp()
{
    int i = 0, j = 0;
    while(i < n)
    {
        if(j==-1 || s[i]==p[j])
        {
            ++i;
            ++j;
        }
        else j = next[j];
    }
    return j;
}

int main()
{
    while(~scanf("%s %s", p, s))
    {
        n = strlen(s);
        m = strlen(p);
        getnext();
        int pos = kmp();
        if(!pos)
        {
            printf("0\n");
            continue;
        }
        for(int i = 0; i < pos; ++i) printf("%c", p[i]);
        printf(" %d\n", pos);
    }
    return 0;
}

猜你喜欢

转载自blog.csdn.net/jay__bryant/article/details/80731550
今日推荐