Palindrome

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 770 Accepted Submission(s): 294

Problem Description

Alice like strings, especially long strings. For each string, she has a special evaluation system to judge how elegant the string is. She defines that a string S[1..3n−2](n≥2) is one-and-half palindromic if and only if it satisfies S[i]=S[2n−i]=S[2n+i−2](1≤i≤n).For example, abcbabc is one-and-half palindromic string, and abccbaabc is not. Now, Alice has generated some long strings. She ask for your help to find how many substrings which is one-and-half palindromic.

Input

The first line is the number of test cases. For each test case, there is only one line containing a string(the length of strings is less than or equal to 500000), this string only consists of lowercase letters.

Output

For each test case, output a integer donating the number of one-and-half palindromic substrings.

Sample Input

1 ababcbabccbaabc

Sample Output

Hint

In the example input, there are two substrings which are one-and-half palindromic strings, $abab$ and $abcbabc$.

题意：

有多少对i,j(i<j)满足S[i-(j-i)...j],S[i...j+(j-i)]为回文串，即以i，j为中心的回文串能够相互覆盖到

做法：

我们设len[i]表示，以i为中心的回文半径可以用马拉车求解；

相互覆盖即： i+len[i]>=j j-len[j]<=i

j-len[j]<=i 可以用树状数组处理。

代码如下：

#include <bits/stdc++.h>
using namespace std;
const int maxn=500005;
typedef long long ll;
vector<int> ve[maxn];
int a[maxn],n;
char s[maxn];
int Len[maxn],tlen[maxn];
int Manacher(char *st,int len){
     int 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]){
            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 lowbit(int x){
    return x&(-x);
}
void add(int pos,int v){
    while(pos<=n){
        a[pos]+=v;
        pos+=lowbit(pos);
    }
}
int query(int pos){
    int ans=0;
    while(pos){
        ans+=a[pos];
        pos-=lowbit(pos);
    }
    return ans;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        memset(a,0,sizeof(a));
        for(int i=0;i<maxn;i++)
            ve[i].clear();
        memset(s,0,sizeof(s));
        scanf("%s",s+1);
        int l=strlen(s+1);
        s[0]='$';
        Manacher(s,l);
        for(int i=2;i<=l;i++){
            tlen[i]=Len[i]-1;
            ve[i-tlen[i]].push_back(i);
            //cout<<i-tlen[i]<<endl;
        }
        ll ans=0;  n=l;
        for(int i=1;i<=n;i++){
            for(int j=0;j<ve[i].size();j++){
                //cout<<ve[i][j]<<endl;
                add(ve[i][j],1);
            }
            //cout<<query(min(i+tlen[i],n))<<" "<<query(i)<<endl;
            ans+=query(min(i+tlen[i],n))-query(i);
        }
        printf("%lld\n",ans);
    }
    return 0;
}

HDU 6230-Palindrome 马拉车算法+树状数组

Palindrome

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