uva 11475 - Extend to Palindrome(KMP)

题目链接：uva 11475 - Extend to Palindrome

题目大意：给定一个字符串，输出最少须要加入多少个字符使得字符串变成回文串。

解题思路：以字符串的转置做KMP，然后用原串匹配就可以。最后匹配长度即为反复长度。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1e5+5;

char s[maxn], t[maxn];
int n, jump[maxn];

void get_jump () {
    int p = 0;
    for (int i = 2; i <= n; i++) {
        while (p && s[p + 1] != s[i])
            p = jump[p];

        if (s[p + 1] == s[i])
            p++;
        jump[i] = p;
    }
}

int find () {
    int p = 0;
    for (int i = 1; i <= n; i++) {
        while (p && s[p + 1] != t[i])
            p = jump[p];

        if (s[p + 1] == t[i])
            p++;
    }
    return p;
}

int main () {
    while (scanf("%s", s + 1) == 1) {
        printf("%s", s+1);
        n = strlen(s + 1);

        for (int i = 1; i <= n + 1; i++)
            t[i] = s[i];

        reverse(s + 1, s + n + 1);
        get_jump();

        int k = find();

        for (int i = k + 1; i <= n; i++)
            printf("%c", s[i]);
        printf("\n");
    }
    return 0;
}