zoukankan      html  css  js  c++  java
  • [POJ2001]Shortest Prefixes

    题目链接:http://poj.org/problem?id=2001

     

    Description

    A prefix of a string is a substring starting at the beginning of the given string. The prefixes of "carbon" are: "c", "ca", "car", "carb", "carbo", and "carbon". Note that the empty string is not considered a prefix in this problem, but every non-empty string is considered to be a prefix of itself. In everyday language, we tend to abbreviate words by prefixes. For example, "carbohydrate" is commonly abbreviated by "carb". In this problem, given a set of words, you will find for each word the shortest prefix that uniquely identifies the word it represents.

    In the sample input below, "carbohydrate" can be abbreviated to "carboh", but it cannot be abbreviated to "carbo" (or anything shorter) because there are other words in the list that begin with "carbo".

    An exact match will override a prefix match. For example, the prefix "car" matches the given word "car" exactly. Therefore, it is understood without ambiguity that "car" is an abbreviation for "car" , not for "carriage" or any of the other words in the list that begins with "car".

    Input

    The input contains at least two, but no more than 1000 lines. Each line contains one word consisting of 1 to 20 lower case letters.

    Output

    The output contains the same number of lines as the input. Each line of the output contains the word from the corresponding line of the input, followed by one blank space, and the shortest prefix that uniquely (without ambiguity) identifies this word.

    Sample Input

    carbohydrate
    cart
    carburetor
    caramel
    caribou
    carbonic
    cartilage
    carbon
    carriage
    carton
    car
    carbonate
    

    Sample Output

    carbohydrate carboh
    cart cart
    carburetor carbu
    caramel cara
    caribou cari
    carbonic carboni
    cartilage carti
    carbon carbon
    carriage carr
    carton carto
    car car
    carbonate carbona

    这个题…真是「哔」了大野狗了,不应该卡这么久的。哎,卡在了下面注释的地方,好歹今天做出来了,否则该睡不了觉了。我可真是水啊……

     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 char str[1010][25];
    16 typedef struct Node {
    17     Node *next[26];
    18     int cnt;
    19     Node() {
    20         cnt = 0;
    21         for(int i = 0; i < 26; i++) {
    22             next[i] = NULL;
    23         }
    24     }
    25 }Node;
    26 
    27 void insert(Node *p, char *str) {
    28     int len = strlen(str);
    29     for(int i = 0; i < len; i++) {
    30         int t = str[i] - 'a';
    31         if(p->next[t] == NULL) {
    32             p->next[t] = new Node();
    33         }
    34         p = p->next[t];
    35         p->cnt++;
    36     }
    37 }
    38 
    39 int find(Node *p, char *str) {
    40     for(int i = 0; str[i]; i++) {
    41         int t = str[i] - 'a';
    42         p = p->next[t];
    43         printf("%c", str[i]);
    44         if(p->cnt == 1) {   //就是这里
    45             break;
    46         }
    47     }
    48 }
    49 
    50 int main() {
    51     int n = 0;
    52     Node *root = new Node();
    53     // freopen("D.in", "r", stdin);
    54     while(gets(str[n]) && strlen(str[n])) {
    55         insert(root, str[n]);
    56         n++;
    57     }
    58     for(int i = 0; i < n; i++) {
    59         printf("%s ", str[i]);
    60         find(root, str[i]);
    61         printf("
    ");
    62     }
    63     return 0;
    64 }
  • 相关阅读:
    socket的一个错误的解释SocketException以及其他几个常见异常
    lambda表达式和ef的语句转化
    多线程的异步委托初识
    跨线程访问控件的方法
    P1337 [JSOI2004]平衡点 / 吊打XXX
    P4099 [HEOI2013]SAO
    UVA10529 Dumb Bones
    P1291 [SHOI2002]百事世界杯之旅
    P2675 《瞿葩的数字游戏》T3-三角圣地
    P4363 [九省联考2018]一双木棋chess
  • 原文地址:https://www.cnblogs.com/kirai/p/4756322.html
Copyright © 2011-2022 走看看