Hdu-5371 Hotaru's problem(马拉车算法)

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 than109 , 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

 

Author

UESTC

 

Source

2015 Multi-University Training Contest 7

 

Recommend

wange2014

分析:我们先用马拉车求出以每个字符间隙向左右延展的回文串最大半长,设第i个间隙的半长为f[i],那么我们要求的其实就是满足存在f[i+x] >= x && f[i] >= x 的最大的x.

我们考虑从左向右枚举i,并将f[i]减i,这样相当于每次询问[i+1,2*i+f[i]]这段区间中所有大于等于i的值的下标的最大值,因为i是单调递增的,所以我们可以把f[i]排序后用线段树在区间上更新下标,复杂度nlogn.

#include<bits/stdc++.h>#define N 100005using namespace std;typedef pair<int,int> pii;pii g[N];int n,m,M,T,Case,ans,s[N],f[N],str[N*2],p[N*2],val[N*4];void manacher(int n){    int i;    for(i = 1;i <= n;i++) str[i*2] = s[i],str[i*2+1] = -1;    str[0] = -2,str[1] = -1;    int up = 2*i;    int res = 0,k = 0,maxk = 0;    for(int i = 2;i < up;i++)    {        p[i] = i < maxk ? min(maxk - i,p[2*k-i]) : 1;        while(str[i-p[i]] == str[i+p[i]]) p[i]++;        if(p[i] + i > maxk) k = i,maxk = i + p[i];    }    for(int i = 3;i < up;i += 2) f[(i+1)/2-1] = (p[i] - 1)/2;}void Update(int p, int v)         //val空间开四倍{    p += M;    val[p] = v;    while(p > 1){        p >>= 1;        val[p] = max(val[p<<1], val[p<<1|1]);    }}int Query(int l, int r) //[l, r]{    int ans = 0;    for(l = l + M - 1, r = r + M + 1; l^r^1; l >>= 1, r >>= 1){        if(~l&1) ans = max(val[l^1], ans); // l为偶数        if(r&1) ans = max(val[r^1], ans); // r为奇数    }    return ans;}int main(){    scanf("%d",&T);    while(T--)    {        ans = 0;        memset(val,0,sizeof(val));        scanf("%d",&n);        for(int i = 1;i <= n;i++) scanf("%d",&s[i]);        manacher(n);        m = n-1;        for(M = 1; M < m + 2; M <<= 1);        for(int i = 1;i < n;i++) g[i] = make_pair(f[i] - i,i);        sort(g+1,g+n);        int tail = n-1;        for(int i = 1;i < n-1;i++)        {            while(tail && (g[tail].first + i) >= 0)            {                Update(g[tail].second,g[tail].second);                tail--;            }            ans = max(ans,Query(i+1,i+f[i]) - i);        }        printf("Case #%d: %d\n",++Case,ans*3);    }}