HDU3068 最长回文
最长回文
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35727 Accepted Submission(s): 13067
Problem Description
给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度.
回文就是正反读都是一样的字符串,如aba, abba等
Input
输入有多组case,不超过120组,每组输入为一行小写英文字符a,b,c...y,z组成的字符串S
两组case之间由空行隔开(该空行不用处理)
字符串长度len <= 110000
Output
每一行一个整数x,对应一组case,表示该组case的字符串中所包含的最长回文长度.
Sample Input
aaaa abab
Sample Output
4 3
这一题第一次写的时候,报Runtime Error(ACCESS_VIOLATION)错误,后来把string str;换成char str[maxn];把字符输入形式由cin>>str换成scanf(“%s”,str)(用cin很容易超时)就AC了;
代码如下:(RE代码)
#include <cstdio> #include <algorithm> #include <cstring> #include <cstdlib> #include <cmath> #include <iostream> #define maxn 110010 using namespace std; int main(){ string str; int len; int maxlen, id; int p[maxn * 2]; int value; bool flag = 0; while (cin>>str){ if (flag) printf("\n"); len = str.length(); for (int i = 2 * len + 1; i>0; i -= 2){ str[i] = '#'; str[i - 1] = str[i / 2 - 1]; } str[0] = '@'; str[2 * len + 2] = '\n'; maxlen = 0; id = 0; value = 0; memset(p, 0, sizeof(p)); for (int i = 1; i<2 * len + 1; i++){ p[i] = maxlen>i ? min(p[2 * id - i], maxlen - i) : 1; while (str[i + p[i]] == str[i - p[i]]) p[i]++; if (maxlen<p[i] + i) { maxlen = p[i] + i; id = i; } if (value < p[i]) value = p[i] - 1; } printf("%d\n", value); } return 0; }
AC代码:
#include <cstdio> #include <algorithm> #include <cstring> #include <cstdlib> #include <cmath> #include <iostream> #define maxn 110010 using namespace std; int main(){ char str[maxn*2]; int len; int maxlen, id; int p[maxn * 2]; int value; bool flag = 0; while (scanf("%s",str)!=EOF){ if (flag) printf("\n"); len = strlen(str); for (int i = 2 * len + 1; i>0; i -= 2){ str[i] = '#'; str[i - 1] = str[i / 2 - 1]; } str[0] = '@'; str[2 * len + 2] = '\0'; maxlen = 0; id = 0; value = 0; memset(p, 0, sizeof(p)); for (int i = 1; i<2 * len + 1; i++){ p[i] = maxlen>i ? min(p[2 * id - i], maxlen - i) : 1; while (str[i + p[i]] == str[i - p[i]]) p[i]++; if (maxlen<p[i] + i) { maxlen = p[i] + i; id = i; } if (value < p[i]) value = p[i] - 1; } printf("%d\n", value); } return 0; }
以下是我觉得讲的不错的关于马拉车算法的博客:
https://blog.csdn.net/WhereIsHeroFrom/article/details/79719507
https://blog.csdn.net/xingyeyongheng/article/details/9310555
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