HDU-2594-Simpsons' Hidden Talents(kmp, 扩展kmp)

链接:

https://vjudge.net/problem/HDU-2594#author=0

题意:

求S1的前缀和S2的后缀的《最大》匹配

思路:

kmp方法:
将s1, s2首尾连接, 根据Next数组求, 注意长度要比s1, 和s2的长度小.
ExtenKmp:
考虑以s1为模板串, 匹配s2, 对于第一个满足exten[i]+i == len的点, 就为最长的长度.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#include <iostream>
#include <sstream>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 1e5+10;

char s1[MAXN], s2[MAXN];
int Next[MAXN];

void GetNext(char *s)
{
    int j = 0;
    int k = -1;
    int len = strlen(s);
    Next[0] = -1;
    while (j < len)
    {
        if (k == -1 || s[j] == s[k])
        {
            ++j;
            ++k;
            Next[j] = k;
        }
        else
            k = Next[k];
    }
}

int main()
{
    while (~scanf("%s%s", s1, s2))
    {
        int lens1 = strlen(s1);
        int lens2 = strlen(s2);
        strcat(s1, s2);
        GetNext(s1);
        int ans = Next[strlen(s1)];
        while (ans > lens1 || ans > lens2)
            ans = Next[ans];
        if (ans == 0 || ans == -1)
            puts("0");
        else
        {
            s1[ans] = 0;
            printf("%s %d
", s1, ans);
        }
    }

    return 0;
}

ExtenKmp

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#include <iostream>
#include <sstream>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 1e5+10;

char s1[MAXN], s2[MAXN];
int Next[MAXN], Exten[MAXN];

void GetNext(char *s)
{
    int len = strlen(s);
    int a = 0, p = 0;
    Next[0] = len;
    for (int i = 1;i < len;i++)
    {
        if (i >= p || i+Next[i-a] >= p)
        {
            if (i >= p)
                p = i;
            while (p < len && s[p] == s[p-i])
                p++;
            Next[i] = p-i;
            a = i;
        }
        else
            Next[i] = Next[i-a];
    }
}

void ExKmp(char *s, char *t)
{
    int len = strlen(s);
    int a = 0, p = 0;
    GetNext(t);
    for (int i = 0;i < len;i++)
    {
        if (i >= p || i + Next[i-a] >= p)
        {
            if (i >= p)
                p = i;
            while (p < len && s[p] == t[p-i])
                p++;
            Exten[i] = p-i;
            a = i;
        }
        else
            Exten[i] = Next[i-a];
    }

}

int main()
{
    while (~scanf("%s%s", s1, s2))
    {
        ExKmp(s2, s1);
        int len = 0, p = 0;
        int lens = strlen(s2);
        for (int i = 0;i < lens;i++)
        {
            if (Exten[i]+i == lens)
            {
                len = Exten[i];
                p = i;
                break;
            }
        }
        if (len == 0)
            puts("0");
        else
            printf("%s %d
", s2+p, len);
    }

    return 0;
}
原文地址:https://www.cnblogs.com/YDDDD/p/11587680.html