Hdu 5340 Three Palindromes maximum input string Manacher,

Hdu 5340 Three Palindromes maximum input string Manacher,
Three PalindromesTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 80 Accepted Submission (s): 21


Problem DescriptionCan we divided a given string S into three nonempty palindromes?
 
InputFirst 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
 
OutputFor each case, output the "Yes" or "No" in a single line.
 
Sample Input

2abcabaadada

 
Sample Output

YesNo

The meaning of the question is to give a string, whether it can be divided into three non-empty return strings. We can find that the first and third strings must be a specific string of the maximum input string. Manacher can find the length of the maximum input string, enumerate the first and last strings, and directly judge in the middle, check whether the maximum input string of the midpoint is included. The complexity is o (n * n). For details about the manacher algorithm, see Manacher algorithm.

#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;}

SourceBestCoder Round #49 ($)
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