【回文子串的个数】HDU1544Palindromes【字符缩距/奇偶判断】

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1544

Problem Description

A regular palindrome is a string of numbers or letters that is the same forward as backward. For example, the string "ABCDEDCBA" is a palindrome because it is the same when the string is read from left to right as when the string is read from right to left.

Now give you a string S, you should count how many palindromes in any consecutive substring of S.

Input

There are several test cases in the input. Each case contains a non-empty string which has no more than 5000 characters.

Proceed to the end of file.

Output

A single line with the number of palindrome substrings for each case.

Sample Input

abaaa

Sample Output

奇偶回文判断AC代码：时间复杂度O(n^2)

#include<iostream>#include<cstring>#include<string>using namespace std;int main(){    string s;    cin.sync_with_stdio(false);    while(cin>>s){        int len=s.size();        int ans=len;        //  长度为1的个数;        for(int i=0;i<len;i++){            int l=i-1,r=i+1;        //  以i为中心的奇数长度回文子串的判断;            while(l>=0&&r<len&&s[l]==s[r]){                l--,r++;                ans++;            }            l=i,r=i+1;              //  以i为中心的偶数长度回文子串的判断;            while(l>=0&&r<=len&&s[l]==s[r]){                l--,r++;                ans++;            }        }        cout<<ans<<endl;    }    return 0;}

字符缩距AC代码：字符串所有字符都不同,最差复杂度O(n^2) 相对于奇偶判断效率还是更高的

#include<iostream>#include<cstring>#include<string>using namespace std;int Hash[5500];int main(){    string s;    cin.sync_with_stdio(false);    while(cin>>s){        memset(Hash,0,sizeof(Hash));        string str="";        str+=s[0];        int k=0;        Hash[0]=1;        for(int i=1;i<s.size();i++){            if(s[i]==s[i-1])                Hash[k]++;            else{                k++;                Hash[k]=1;                str+=s[i];            }        }        k++;        int ans=0;        for(int i=0;i<k;i++)            ans+=(1+Hash[i])*Hash[i]/2;        for(int i=0;i<k;i++){            int a=i-1;            int b=i+1;            if(a<0||b>=k) continue;     //  一言不合就跳过            int flag=0;            while(str[a]==str[b]&&Hash[a]==Hash[b]){                ans+=Hash[a];                a--;                b++;                if(a<0||b>=k){                    flag=1;                    break;                }            }            if(flag==0&&str[a]==str[b]){                ans+=min(Hash[a],Hash[b]);            }        }        cout<<ans<<endl;    }    return 0;}

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