UVA 11475 - Extend to Palindrome(KMP)

UVA 11475 - Extend to Palindrome

题目链接

题意：给定一个字符串，问需要补上最少的字符使他变成回文串

思路：KMP，把字符串逆序和原串做匹配，匹配到最后一个字符看匹配了多少个，就是最大重合部分，然后相应输出即可

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXLEN = 100005;struct KMP {    int next[MAXLEN];    void getnext(char *str) {int n = strlen(str);next[1] = 0;int j = 0;for (int i = 2; i <= n; i++) {    while (j && str[j] != str[i - 1]) j = next[j];    if (str[j] == str[i - 1]) j++;    next[i] = j;}    }    void find(char *T, char *s) {int n = strlen(T);getnext(s);int j = 0;for (int i = 0; i < n; i++) {    while (j && T[i] != s[j]) j = next[j];    if (T[i] == s[j]) j++;}printf("%s%s\n", T, s + j);    }};KMP gao;char a[MAXLEN], b[MAXLEN];int main() {    while (~scanf("%s", a)) {strcpy(b, a);reverse(b, b + strlen(b));gao.find(a, b);    }    return 0;}