HDU 3068 ( 最长回文 )

Problem : 3068 ( 最长回文 )     Judge Status : AcceptedRunId : 6255711    Language : C++    Author : ssunCode Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta#include "stdio.h"#include "string.h"#define N 2200050char str[N],nstr[2*N];int rad[2*N];int len,nlen;int manacher(){    int id, i, ans = 1;    int mx = 0;//    printf("%d",strlen(nstr));    for(i=1; i<nlen; i++){//$#1#2#3#3#2#1#不要用strlen(nstr)代替nlen        if(mx > i)            rad[i] = mx - i < rad[2*id-i] ? mx - i : rad[2*id-i];        else            rad[i] = 1;        for(;nstr[i+rad[i]] == nstr[i-rad[i]];rad[i]++)            ;        if(rad[i] + i > mx){            mx = rad[i] + i;            id = i;        }        ans = rad[i] > ans ? rad[i] : ans;    }    return ans-1;}int main(){    while(scanf("%s",str)!=EOF){        int i=0;//        memset(rad,0,sizeof(rad));        nstr[0] = '$';        nstr[1] = '#';        len = strlen(str);        for(i=0; i<len; i++){            nstr[2*i+3] = '#';            nstr[2*i+2] = str[i];                    }        nstr[2*i+2] = 0;        nlen = 2 * len + 2;        printf("%d\n",manacher());    }    return 0;}

参考http://bbs.dlut.edu.cn/bbstcon.php?board=Competition&gid=23474对Manacher算法的解释

其实值得一提的是这个程序的manacher()函数里面的for循环不要用strlen()函数，不然的话会超时，不用strlen函数而用nlen = 2 * len + 2的话250ms可以过