HDU 3068 ( 最长回文 )

Problem : 3068 ( 最长回文 )     Judge Status : Accepted
RunId : 6255711    Language : C++    Author : ssun
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta
#include "stdio.h"
#include "string.h"
#define N 2200050

char 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