HDOJ 5371 Hotaru's problem manacher+优先队列+二分

先用求回文串的Manacher算法，求出以第i个点和第i+1个点为中心的回文串长度，记录到数组c中 比如 10 9 8 8 9 10 10 9 8 我们通过运行Manacher求出第i个点和第i+1个点为中心的回文串长度 0 0 6 0 0 6 0 0 0

两个8为中心，10 9 8 8 9 10是个回文串，长度是6。 两个10为中心，8 9 10 10 9 8是个回文串，长度是6。

要满足题目所要求的内容，需要使得两个相邻的回文串，共享中间的一部分，比如上边的两个字符串，共享 8 9 10这一部分。 也就是说，左边的回文串长度的一半，要大于等于共享部分的长度，右边回文串也是一样。 因为我们已经记录下来以第i个点和第i+1个点为中心的回文串长度， 那么问题可以转化成，相距x的两个数a[i],a[i+x]，满足a[i]/2>=x 并且 a[i+x]/2>=x，要求x尽量大

这可以用一个set维护，一开始集合为空，依次取出a数组中最大的元素，将其下标放入set中，每取出一个元素，再该集合中二分查找比i+a[i]/2小，但最大的元素，更新答案。 然后查找集合中比i-a[i]/2大，但最小的元素，更新答案。

答案就是3*an

Hotaru's problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1513    Accepted Submission(s): 555

Problem Description

Hotaru Ichijou recently is addicated to math problems. Now she is playing with N-sequence.
Let's define N-sequence, which is composed with three parts and satisfied with the following condition:
1. the first part is the same as the thrid part,
2. the first part and the second part are symmetrical.
for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.

Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.

 

Input

There are multiple test cases. The first line of input contains an integer T（T<=20）, indicating the number of test cases. 

For each test case:

the first line of input contains a positive integer N（1<=N<=100000）, the length of a given sequence

the second line includes N non-negative integers ,each interger is no larger than 109 , descripting a sequence.

 

Output

Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by a integer, the largest length of N-sequence.

We guarantee that the sum of all answers is less than 800000.

 

Sample Input

1102 3 4 4 3 2 2 3 4 4

 

Sample Output

Case #1: 9

 

Source

2015 Multi-University Training Contest 7

 

/* ***********************************************Author        :CKbossCreated Time  :2015年08月12日 星期三 12时44分58秒File Name     :HDOJ5371.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <set>using namespace std;const int maxn=100100;int n;int str[maxn],ans[maxn*3];int p[maxn*3],pos,tot;void pre() { tot=1;memset(ans,0,sizeof(ans));ans[0]=-2;for(int i=0;i<n;i++){ans[tot]=-1; tot++;ans[tot]=str[i]; tot++;}ans[tot]=-1;tot++;}void manacher(){pos=-1;memset(p,0,sizeof(p));int mx=-1,mid=-1;int len=tot;for(int i=0;i<len;i++){int j=-1;if(i<mx){j=2*mid-i;p[i]=min(p[j],mx-i);}else p[i]=1;while(i+p[i]<len&&ans[i+p[i]]==ans[i-p[i]])p[i]++;if(p[i]+i>mx){mx=p[i]+i; mid=i;}}}struct Node{int pos,len;Node(){}Node(int _pos,int _len):pos(_pos),len(_len){}void toString(){printf("pos: %d len: %d\n",pos,len);}bool operator<(const Node& node) const{return len<node.len;}};int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);int cas=1,T_T;scanf("%d",&T_T);while(T_T--){scanf("%d",&n);for(int i=0;i<n;i++)scanf("%d",str+i);pre();manacher();priority_queue<Node> q;for(int i=0;i<tot;i++){if(ans[i]==-1){Node node(i/2,p[i]/2);q.push(node);}}set<int> st;set<int>::iterator it;int as=0;while(!q.empty()){Node u=q.top(); q.pop();if(st.size()==0) { st.insert(u.pos); continue; }int left=u.pos-u.len;it=st.lower_bound(left);if(*it<left||it==st.end()) it--;if(*it>=left){as=max(as,u.pos-*it);}int right=u.pos+u.len;it=st.lower_bound(right);if(*it>right||it==st.end()) it--;if(*it<=right){as=max(as,*it-u.pos);}st.insert(u.pos);}printf("Case #%d: %d\n",cas++,as*3);}        return 0;}