最长回文
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17329 Accepted Submission(s): 6377
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
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
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
aaaaabab
Sample Output
4 3
change数组和p数组要记得开二倍，忘记开的话会WA
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;const int maxn = 120005;int p[2*maxn];
char str[2*maxn];
char strsc[maxn];void init()
{str[0] = '@';str[1] = '#';int len = strlen(strsc);for(int i=0;i<len;i++){str[2*i+2] = strsc[i];str[2*i+3] = '#';}str[2*len+2] = '\0';
}int manacher()
{int ans = -1;int mx = 0,id = 0;int len = strlen(str);for(int i=0;i<len;i++){if(i > id)p[i] = min(p[2*id-i],mx-i);elsep[i] = 1;while(str[i+p[i]] == str[i-p[i]])p[i]++;if(i + p[i] > mx){mx = i + p[i];id = i;}ans = max(ans,p[i]);}return ans - 1;
}int main()
{while(~scanf("%s",&strsc)){init();printf("%d\n",manacher());}
}