HDU5384-Hotaru's problem-Manacher

找出紧挨的三个回文串，例如abccbaabc ，形如ABA格式，其中AB为回文串。计算最长的长度。

首先用Manacher处理回文半径。然后就是找到两个点，都是偶数的回文串，并且共享了中间一段。

之后拿set搞一下就可以了= =

#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn = 1e5+10;
int Ma[maxn<<2],Mp[maxn<<2];
set <int> pos;

struct node{
    int mp,p;
    node(){}
    bool operator < (const node &rhs) const
    {
        if(mp == rhs.mp) return p > rhs.p;
        else return mp < rhs.mp;
    }
};

priority_queue<node> pq;
void Manacher(int s[],int len)
{
    int l = 0;
    Ma[l++] = '$';
    Ma[l++] = '#';
    for(int i=0;i<len;i++)
    {
        Ma[l++] = s[i];
        Ma[l++] = '#';
    }
    Ma[l] = 0;
    int mx = 0,id = 0;
    for(int i=0;i<l;i++)
    {
        Mp[i] = mx > i? min(Mp[2*id-i],mx-i) : 1;
        while(Ma[i + Mp[i]] == Ma[i-Mp[i]]) Mp[i]++;
        if(Mp[i] + i > mx)
        {
            mx = Mp[i] + i;
            id = i;
        }
    }
}

int T,N;
int save[maxn];
int main()
{
    scanf("%d",&T);
    int cas = 0;
    while(T--)
    {
        scanf("%d",&N);
        for(int i=0;i<N;i++) scanf("%d",&save[i]);
        Manacher(save,N);
        int ans = 0;
        node tmp;
        while(!pq.empty()) pq.pop();
        pos.clear();
        for(int i=0;i<2*N+1;i++)
        {
            if(Ma[i] == '#' && Mp[i] > 1)
            {
                tmp.mp = Mp[i];
                tmp.p = i;
                pq.push(tmp);
            }
        }
        int cnt = 0;
        //for(int i=0;i<2*N+1;i++) printf("%d	",i); puts("");
        //for(int i=0;i<2*N+1;i++) printf("%c	",Ma[i]=='#'?'#':Ma[i]+'0'); puts("");
        //for(int i=0;i<2*N+1;i++) printf("%d	",Mp[i]); puts("");
        while(!pq.empty())
        {
            tmp = pq.top(); pq.pop();
            int cur = tmp.p,mp = tmp.mp;
            //printf("cur:%d mp:%d
",cur,mp);
            int tmp_pos = 0;
            if(cnt)
            {
                auto it = pos.lower_bound(cur - (mp-1) );

                //printf("L:%d ",*it);
                if(it != pos.end() && *it >= cur-(mp-1)) ans = max(ans,cur-*it);

                it = pos.lower_bound(cur + (mp-1) ); it--;
                //printf("R:%d
",*it);
                if(it != pos.end() && *it <= cur+(mp-1)) ans = max(ans,*it-cur);
            }
            pos.insert(cur);
            cnt++;
        }
        printf("Case #%d: %d
",++cas,ans/2*3);
    }
}