Hdu 5340 Three Palindromes 最大回文串 Manacher

Three Palindromes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 80    Accepted Submission(s): 21

Problem Description

Can we divided a given string S into three nonempty palindromes?

 

Input

First line contains a single integer T≤20 which denotes the number of test cases.

For each test case , there is an single line contains a string S which only consist of lowercase English letters.1≤|s|≤20000

 

Output

For each case, output the "Yes" or "No" in a single line.

 

Sample Input

2abcabaadada

 

Sample Output

YesNo

 题意，是给出一个字符串，能否分成三个非空回文串。

我们可以发现 第一个和第三个串，一定是最大回文串的某个串,Manacher 求出最大回文串的长度，枚举第一个和最后一个，中间直接判断，中点的最大回文串是否包括了就可以了。复杂度为o(n * n).

具体manacher算法参见 Manacher算法

#define N 110050#define M 100005#define maxn 205#define MOD 1000000000000000007int T,n,a,pri[N],ans,len,sn = 0,top[N],tail[N],pn,ln;bool dp[N][4];pii seg[N];bool Manacher(char str[],int len){    char tstr[N+N];    int p[N + N],l2 =0,mi;    tstr[l2++] = '#';    for(int i =0;i<len;i++){        tstr[l2++] = str[i];        tstr[l2++] = '#';    }    p[0] = 0;mi = 0;    for(int i = 1;i<l2;i++){        int mi2 = mi + mi - i;        if(mi + p[mi] >= i) p[i] = min(mi2 - (mi - p[mi]),p[mi2]);        else p[i] = 0;        if(p[i] == 0 ||  mi2 - p[mi2] == mi - p[mi]){            int maxx = p[i]+1;            while(i- maxx >= 0 && i+maxx < l2 && tstr[i-maxx] == tstr[i+maxx]){                maxx++;            }            p[i] = maxx - 1;        }        if(p[i] + i > p[mi] + mi) mi = i;    }    int ans = -1;sn = 0;pn = ln = 0;    for(int i = 1;i < l2 - 1;i++){        if(i - p[i] == 0) top[pn++]  = i;        if(i + p[i] == l2 - 1) tail[ln++] = i;    }    for(int i = 0;i < pn;i++){        for(int j = ln - 1;j>=0;j--){            int s1 = top[i] + p[top[i]] + 1,s2 = tail[j] - p[tail[j]] - 1;            if(s1 > s2 )            break;            int mid = (s1 + s2)/2;            if(p[mid] >= mid - s1) return true;        }    }    return false;    //printf("%d\n",ans);}char str[N];int main(){    while(S(T)!=EOF)    {        while(T--){            SS(str);            len = strlen(str);            if(Manacher(str,len))                printf("Yes\n");            else                printf("No\n");        }    }    return 0;}

Source

BestCoder Round #49 ($)

 

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