Girls' research（hdu3294+Manacher算法）

Girls' research

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1027    Accepted Submission(s): 389

Problem Description

 

One day, sailormoon girls are so delighted that they intend to research about palindromic strings. Operation contains two steps:
First step: girls will write a long string (only contains lower case) on the paper. For example, "abcde", but 'a' inside is not the real 'a', that means if we define the 'b' is the real 'a', then we can infer that 'c' is the real 'b', 'd' is the real 'c' ……, 'a' is the real 'z'. According to this, string "abcde" changes to "bcdef".
Second step: girls will find out the longest palindromic string in the given string, the length of palindromic string must be equal or more than 2.

 

 

Input

 

Input contains multiple cases.
Each case contains two parts, a character and a string, they are separated by one space, the character representing the real 'a' is and the length of the string will not exceed 200000.All input must be lowercase.
If the length of string is len, it is marked from 0 to len-1.

 

 

Output

 

Please execute the operation following the two steps.
If you find one, output the start position and end position of palindromic string in a line, next line output the real palindromic string, or output "No solution!".
If there are several answers available, please choose the string which first appears.

 

 

Sample Input

 

b babd a abcd

 

 

Sample Output

 

0 2 aza No solution!

 

 

Author

 

wangjing1111

 

 

Source

 

2010 “HDU-Sailormoon” Programming Contest

 

 

 

 

题意：求最长回文子串，并求出起始位置。
分析：预处理移位表，之后用Manacher算法搞定

 

转载请注明出处：寻找&星空の孩子  

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3294

 

 

//Manacher算法
#include
#include
#include
using namespace std;
#define maxn 200010*2

int P[maxn];
//(p.s. 可以看出，P[i]-1正好是原字符串中回文串的总长度）
char s1[maxn];
char s2[maxn];
char ch[30];

void manacher(char* s)
{
    int i,id=0,mx=0;
    P[0]=0;                //P[0]位置没用
    for(i=1;s[i];i++)      //对串进行线性扫描
    {
        if(mx > i)         //如果mx比当前i大，分为两种情况，详细致请看文章开头推荐的blog上的图示，非常给力的图
            P[i] = min(P[2*id-i],mx-i);
        else               //如果mx比i小，没有可以利用的信息，那么就只能从头开始匹配
            P[i] = 1;
        while(s[i+P[i]]==s[i-P[i]] )P[i]++;     //匹配
        if(mx < P[i] + i) //坚持是否有更新mx以及id
        {
            mx = P[i] + i;
            id = i;
        }
    }
}


void init(int t)
{
    int i, j = 2;
    s2[0] = '$', s2[1] = '#';
    for(i=0;s1[i];i++)
    {
        s2[j++] = s1[i];
        s2[j++] = '#';
    }
    s2[j] = '\0';
    //置换表
    char ex='a';
    ch[t]=ex++;
    for(i=t+1;(i%26)!=t;i++)
    {
        ch[i%26]=ex++;
    }
//    for(i=0;i<=25;i++)
//    printf("%c",ch[i]);
}

int main()
{
    char gh,gg;
    while(scanf("%c %s%c",&gh,s1,&gg)!=EOF)
    {
      //  getchar();
        init(gh-'a');
        manacher(s2);
        int ans=0,x,y;
        for(int i=1;s2[i]!='\0';i++)
        {
            if(P[i]>ans)
            {
                ans=P[i];
                x=i;
            }
        }
        //printf("%d\n",ans-1);
        if(ans-1==1) {printf("No solution!\n");continue;}
 //       printf("ans=%d\tx=%d\n",ans,x);
 //       if(x&1){x=(x-1)/2;x=x-(ans-1)/2;y=x+ans-2;}
 //       else {x=(x-1)/2;x=x-(ans-1)/2;y=x+ans-2;}
        //对应起始终点位置
        x=(x-1)/2;
        x=x-(ans-1)/2;
        y=x+ans-2;
        printf("%d %d\n",x,y);
        for(int i=x;i<=y;i++)
        {
            printf("%c",ch[s1[i]-'a']);
        }
        printf("\n");
    }
    return 0;
}
/*
a abbab
a abbabcba
b babd
a abcd
*/