HDU 3294 - Girls' research 马拉车算法模板

Girls' research Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 4365    Accepted Submission(s): 1633   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!

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题意：要你找出最长回文串的开头和结尾，并且要遵守如下规则将最长公共子串变形输出：开始会给你一个即将代替a的的字符，如样例中的b，变成a，则c就是b,....a就是z。

做法：没什么好解释的就是板子套套的事情，注意中间的转化就好。

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代码如下：

#include<bits/stdc++.h>
using namespace std;

const int maxn=200010;
char str[maxn];//原字符串
char tmp[maxn<<1];//转换后的字符串
int Len[maxn<<1],mx,po,ans,ansst,ansen,anslen;
char fi[5],s[maxn];
int init(char *st){//转换原始串
    int i,len=strlen(st);
    tmp[0]='@';//字符串开头增加一个特殊字符，防止越界
    for(i=1;i<=2*len;i+=2){
        tmp[i]='#';
        tmp[i+1]=st[i/2];
    }
    tmp[2*len+1]='#';
    tmp[2*len+2]='$';//字符串结尾加一个字符，防止越界
    tmp[2*len+3]=0;
    return 2*len+1;//返回转换字符串的长度
}//Manacher算法计算过程
int Manacher(char *st,int len){
     mx=0,ans=0,po=0;//mx即为当前计算回文串最右边字符的最大值
     for(int i=1;i<=len;i++){
         if(mx>i)
            Len[i]=min(mx-i,Len[2*po-i]);//在Len[j]和mx-i中取个小
         else
            Len[i]=1;//如果i>=mx，要从头开始匹配
         while(st[i-Len[i]]==st[i+Len[i]])
            Len[i]++;
         if(Len[i]+i>mx){//若新计算的   回文串右端点位置大于mx，要更新po和mx的值
             mx=Len[i]+i;
             po=i;
         }
         if(ans<Len[i]){
            if(i%2==0) ansst=i/2-(Len[i]-1)/2,ansen=i/2+(Len[i]-1)/2;
            else { ansst=(i-1)/2-(Len[i]-1)/2+1,ansen=(i-1)/2+(Len[i]-1)/2;}
            anslen=ansen-ansst;
            ans=Len[i];
         }
     }
     return ans-1;//返回Len[i]中的最大值-1即为原串的最长回文子串额长度
}
int main(){
    while(~scanf("%s%s",fi,s)){
        int len=init(s);
        int ans=Manacher(tmp,len);
        if(ans==1) printf("No solution!\n");
        else {
            ansst--,ansen--;
            int cha=fi[0]-'a';
            printf("%d %d\n",ansst,ansen);
            for(int i=ansst;i<=ansen;i++){
                printf("%c",(s[i]-'a'+26-cha)%26+'a');
            }
            cout<<endl;
        }
    }
    return 0;
}