hdu5371 Hotaru's problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2189    Accepted Submission(s): 774


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
1 10 2 3 4 4 3 2 2 3 4 4
 

Sample Output
Case #1: 9

这题可以用Manacher算法做,因为题目要找的是三段(第一段和第二段对称,第二段和第三段对称),其实就是两个连在一起的回文串,我们可以先用Manacher算法初始化各个点的p[i]值(即可以向右延伸的最大距离,包括本身,这时已经加入了-1代替算法中的'#',-2代替算法中的'$'),然后对于每个i,枚举j(j属于1~p[i]-1),如果i+j-p[i+j]+1<=i,那么说明i,j可以分别作为第一、二段的点和第二、三段的点)。

这里有个优化,因为枚举时满足条件的只有'#'(即'-1’),所以我们可以使i,j每次变化2.

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 100060
int a[maxn],b[2*maxn],p[2*maxn];
int main()
{
    int n,m,i,j,T,mx,idx,maxx,num1=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        if(n<3){
            printf("0\n");continue;
        }
        b[0]=-2;
        b[1]=-1;
        for(i=0;i<n;i++){
            b[i*2+2]=a[i];
            b[i*2+3]=-1;
        }
        n=2*n+2;mx=0;
        for(i=0;i<n;i++){
            if(i<mx){
                p[i]=min(p[idx*2-i],mx-i);
            }
            else p[i]=1;
            while(b[i-p[i]]==b[i+p[i]]){
                p[i]++;
            }
            if(mx<i+p[i]){
                mx=i+p[i];
                idx=i;
            }
        }
        maxx=0;
        for(i=3;i<n;i+=2){
            for(j=p[i]-1;j>=1;j-=2){
                if(j<maxx)break;
                if(i+j-p[i+j]+1<=i){
                    maxx=max(maxx,j);break;
                }
            }

        }
        num1++;
        printf("Case #%d: ",num1);
        printf("%d\n",maxx/2*3);
    }
    return 0;
}


posted @ 2015-08-19 21:29  Herumw  阅读(70)  评论(0编辑  收藏  举报