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  • HDU 5421 Victor and String

    Victor and String

    Time Limit: 1000ms
    Memory Limit: 262144KB
    This problem will be judged on HDU. Original ID: 5421
    64-bit integer IO format: %I64d      Java class name: Main

    Victor loves to play with string. He thinks a string is charming as the string is a palindromic string.

    Victor wants to play n times. Each time he will do one of following four operations.

    Operation 1 : add a char c to the beginning of the string.

    Operation 2 : add a char c to the end of the string.

    Operation 3 : ask the number of different charming substrings.

    Operation 4 : ask the number of charming substrings, the same substrings which starts in different location has to be counted.

    At the beginning, Victor has an empty string.

    Input
    The input contains several test cases, at most 5 cases.

    In each case, the first line has one integer n means the number of operations.

    The first number of next n line is the integer op, meaning the type of operation. If op=1 or 2, there will be a lowercase English letters followed.

    1≤n≤100000.

    Output
    For each query operation(operation 3 or 4), print the correct answer.

    Sample Input

    6
    1 a
    1 b
    2 a
    2 c
    3
    4
    8
    1 a
    2 a
    2 a
    1 a
    3
    1 b
    3
    4

    Sample Output

    4
    5
    4
    5
    11

    Source

     
    解题:Palindromic Tree
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 using LL = long long;
     4 const int maxn = 200010;
     5 struct PalindromicTree {
     6     int son[maxn][26],fail[maxn],num[maxn],s[maxn],len[maxn];
     7     int tot,L,R,last[2];
     8     LL ret;
     9     int newnode(int slen = 0) {
    10         num[tot] = 0;
    11         memset(son[tot],0,sizeof son[tot]);
    12         len[tot] = slen;
    13         return tot++;
    14     }
    15     void init() {
    16         ret = tot = last[0] = last[1] = 0;
    17         L = maxn>>1;
    18         R = L - 1;
    19         newnode(0);
    20         newnode(-1);
    21         fail[0] = fail[1] = 1;
    22         memset(s,-1,sizeof s);
    23     }
    24     int getFail(int kd,int x) {
    25         if(kd) while(s[R - len[x] -1] != s[R]) x = fail[x];
    26         else while(s[L + len[x] + 1] != s[L]) x = fail[x];
    27         return x;
    28     }
    29     void extend(int kd,int c) {
    30         if(kd) s[++R] = c;
    31         else s[--L] = c;
    32         int cur = getFail(kd,last[kd]);
    33         if(!son[cur][c]) {
    34             int x = newnode(len[cur] + 2);
    35             fail[x] = son[getFail(kd,fail[cur])][c];
    36             son[cur][c] = x;
    37             num[x] = num[fail[x]] + 1;
    38         }
    39         last[kd] = son[cur][c];
    40         if(len[last[kd]] == R - L + 1) last[kd^1] = last[kd];
    41         ret += num[last[kd]];
    42     }
    43 } pt;
    44 int main() {
    45     int n,op;
    46     char str[10];
    47     while(~scanf("%d",&n)) {
    48         pt.init();
    49         while(n--) {
    50             scanf("%d",&op);
    51             if(op <= 2) {
    52                 scanf("%s",str);
    53                 pt.extend(op-1,str[0] - 'a');
    54             } else if(op == 3) printf("%d
    ",pt.tot - 2);
    55             else if(op == 4) printf("%I64d
    ",pt.ret);
    56         }
    57     }
    58     return 0;
    59 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4914193.html
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