hdu 5340 （manacher）

Sample Input

2 abc abaadada

 

Sample Output

Yes No

判断是否能成为3个非空回文子串

manacher算法求出个点回文长度，在找出第一个和最后一个保存下来，再判断中间的

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MOD 3221225473
#define MAXN 20005
#define MIN 0
#define MAX 1000001

int n;
char d[MAXN];
char st[MAXN*2];
int p[MAXN*2],begi[MAXN*2],tail[MAXN*2];
int len;
void manacher()
{
    int MaxId=0,id;
    for(int i=0; i<len; i++)
    {
        if(MaxId>i)
            p[i]=min(p[2*id-i],MaxId-i);
        else
            p[i]=1;
        while(st[i+p[i]]==st[i-p[i]])
            p[i]++;

        if(p[i]+i>MaxId)
        {
            id=i;
            MaxId=p[i]+i;
        }
        p[i] -= 1;
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        getchar();
        scanf("%s",d);
        int l = strlen(d);

        len = 0;
        st[len++] = '#';
        for(int i = 0; i < l; i++)
        {
            st[len++] = d[i];
            st[len++] = '#';
        }
        st[len] = 0;
        manacher();

        int flag = 0;
        int pn =0 ,ln = 0;
        for(int i = 1; i < len - 1; i++)
        {
            if(i - p[i] == 0) begi[pn++]  = i;
            if(i + p[i] == len-1) tail[ln++] = i;
        }
        for(int i = 0; i < pn; i++)
        {
            for(int j = ln - 1; j>=0; j--)
            {
                int s1 = begi[i] + p[begi[i]]+1 ,s2 = tail[j] - p[tail[j]]-1;
                if(s1 > s2)
                    break;
                int mid = (s1 + s2)/2;
                if(p[mid] >= mid-s1)
                {
                    flag = 1;
                    break;
                }
            }
            if(flag )
                break;
        }
        if(flag )
            printf("Yes
");
        else
            printf("No
");
    }
    return 0;
}