zoukankan      html  css  js  c++  java
  • [POJ3974]Palindrome

    题目链接:http://poj.org/problem?id=3974

    Description

    Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient algorithm to find the length of the largest palindrome in a string?"

    A string is said to be a palindrome if it reads the same both forwards and backwards, for example "madam" is a palindrome while "acm" is not.

    The students recognized that this is a classical problem but couldn't come up with a solution better than iterating over all substrings and checking whether they are palindrome or not, obviously this algorithm is not efficient at all, after a while Andy raised his hand and said "Okay, I've a better algorithm" and before he starts to explain his idea he stopped for a moment and then said "Well, I've an even better algorithm!".

    If you think you know Andy's final solution then prove it! Given a string of at most 1000000 characters find and print the length of the largest palindrome inside this string.

    Input

    Your program will be tested on at most 30 test cases, each test case is given as a string of at most 1000000 lowercase characters on a line by itself. The input is terminated by a line that starts with the string "END" (quotes for clarity).

    Output

    For each test case in the input print the test case number and the length of the largest palindrome.

    Sample Input

    abcbabcbabcba
    abacacbaaaab
    END

    Sample Output

    Case 1: 13
    Case 2: 6

    Source

     
     
    Manacher,数组要开大一些,否则会有奇怪的RE
     
     
     
     1 #include <cstdio>
     2 #include <cstdlib>
     3 #include <cstring>
     4 #include <algorithm>
     5 #include <iostream>
     6 #include <cmath>
     7 #include <queue>
     8 #include <map>
     9 #include <stack>
    10 #include <list>
    11 #include <vector>
    12 
    13 using namespace std;
    14 
    15 const int maxn = 2222222;
    16 int pre[maxn];
    17 char str[maxn];
    18 char tmp[maxn];
    19 
    20 inline int min(int x, int y) {
    21     return x < y ? x : y;
    22 }
    23 
    24 int init(char *tmp, char *str) {
    25     int len = strlen(str);
    26     tmp[0] = '$';
    27     for(int i = 0; i <= len; i++) {
    28         tmp[2*i+1] = '#';
    29         tmp[2*i+2] = str[i];
    30     }
    31     len = 2 * len + 2;
    32     tmp[len] = 0;
    33     return len;
    34 }
    35 
    36 void manacher(int *pre, char *tmp, int len) {
    37     int id = 0;
    38     int mx = 0;
    39     for(int i = 1; i < len; i++) {
    40         pre[i] = mx > i ? min(pre[2*id-i], mx-i) : 1;
    41         while(tmp[i+pre[i]] == tmp[i-pre[i]]) {
    42             pre[i]++;
    43         }
    44         if(pre[i] + i > mx) {
    45             id = i;
    46             mx = pre[i] + id;
    47         }
    48     }
    49 }
    50 
    51 int main() {
    52     int kase = 1;
    53     // freopen("in", "r", stdin);
    54     while(~scanf("%s", str) && strcmp(str, "END") != 0) {
    55         memset(pre, 0, sizeof(pre));
    56         memset(tmp, 0, sizeof(tmp));
    57         int len = init(tmp, str);
    58         manacher(pre, tmp, len);
    59         int ans = 1;
    60         for(int i = 0; i < len; i++) {
    61             ans = max(ans, pre[i]);
    62         }
    63         printf("Case %d: %d
    ", kase++, ans - 1);
    64     }
    65 }
  • 相关阅读:
    第一次作业 —— 【作业7】问卷调查
    讲座观后感
    学习进度表(随缘更新)
    数据结构与算法思维导图
    作业七问卷调查
    《创新者的逆袭,用第一性原理做颠覆式创新》读后感
    结对项目--四则运算生成器(Java) 刘彦享+龙俊健
    个人项目---WordCount实现(Java)
    自我介绍+软工五问
    简洁又快速地处理集合——Java8 Stream(下)
  • 原文地址:https://www.cnblogs.com/kirai/p/4769191.html
Copyright © 2011-2022 走看看