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  • Codeforces Round #290 (Div. 2) 拓扑排序

    C. Fox And Names
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in the lexicographical order.

    After checking some examples, she found out that sometimes it wasn't true. On some papers authors' names weren't sorted in lexicographical order in normal sense. But it was always true that after some modification of the order of letters in alphabet, the order of authors becomes lexicographical!

    She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the lexicographical order. If so, you should find out any such order.

    Lexicographical order is defined in following way. When we compare s and t, first we find the leftmost position with differing characters: si ≠ ti. If there is no such position (i. e. s is a prefix of t or vice versa) the shortest string is less. Otherwise, we compare characters si and tiaccording to their order in alphabet.

    Input

    The first line contains an integer n (1 ≤ n ≤ 100): number of names.

    Each of the following n lines contain one string namei (1 ≤ |namei| ≤ 100), the i-th name. Each name contains only lowercase Latin letters. All names are different.

    Output

    If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters 'a'–'z' (i. e. first output the first letter of the modified alphabet, then the second, and so on).

    Otherwise output a single word "Impossible" (without quotes).

    Examples
    input
    3
    rivest
    shamir
    adleman
    output
    bcdefghijklmnopqrsatuvwxyz
    input
    10
    tourist
    petr
    wjmzbmr
    yeputons
    vepifanov
    scottwu
    oooooooooooooooo
    subscriber
    rowdark
    tankengineer
    output
    Impossible
    input
    10
    petr
    egor
    endagorion
    feferivan
    ilovetanyaromanova
    kostka
    dmitriyh
    maratsnowbear
    bredorjaguarturnik
    cgyforever
    output
    aghjlnopefikdmbcqrstuvwxyz
    input
    7
    car
    care
    careful
    carefully
    becarefuldontforgetsomething
    otherwiseyouwillbehacked
    goodluck
    output
    acbdefhijklmnogpqrstuvwxyz

    题意:给你n个字符串 现在要求你重新定义字典序 使得n个字符串的顺序为升序 输出新的26个字母的字典序
    题解:对相邻的两个字符串 在第一个字符不同的位置的两个字符间连一条单向边 拓扑排序
     1 #pragma comment(linker, "/STACK:102400000,102400000")
     2 #include <bits/stdc++.h>
     3 #include <cstdlib>
     4 #include <cstdio>
     5 #include <iostream>
     6 #include <cstdlib>
     7 #include <cstring>
     8 #include <algorithm>
     9 #include <cmath>
    10 #include <cctype>
    11 #include <map>
    12 #include <set>
    13 #include <queue>
    14 #include <bitset>
    15 #include <string>
    16 #include <complex>
    17 #define LL long long
    18 #define mod 1000000007
    19 using namespace std;
    20 char a[105][1003];
    21 int g[26][26];
    22 int in[26];
    23 queue<int> q;
    24 int vis[26];
    25 int ans[26];
    26 int jishu=0;
    27 int n;
    28 bool topsort(){
    29     while(!q.empty())
    30         q.pop();
    31     for(int i=0;i<=25;i++){
    32         if(in[i]==0){
    33         q.push(i);
    34         vis[i]=1;
    35         }
    36     }
    37     jishu=0;
    38     int exm;
    39     while(!q.empty()){
    40         exm=q.front();
    41         ans[jishu++]=exm;
    42         q.pop();
    43         for(int i=0;i<=25;i++){
    44             if(g[exm][i]==1){
    45                 if(--in[i]==0&&vis[i]==0){
    46                     vis[i]=1;
    47                     q.push(i);
    48                 }
    49             }
    50         }
    51     }
    52 }
    53 int main()
    54 {
    55     scanf("%d",&n);
    56     for(int i=0;i<n;i++)
    57         scanf("%s",a[i]);
    58     for(int i=0;i<n-1;i++)
    59     {
    60         int l1,l2;
    61         l1=strlen(a[i]);
    62         l2=strlen(a[i+1]);
    63         int p=0;
    64         while(p<min(l1,l2)&&a[i][p]==a[i+1][p]) p++;
    65         if(p==l1&&l1<l2) continue;
    66         if(p==l2&&l2<l1) {
    67             printf("Impossible
    ");//**
    68             return 0;
    69         }
    70         if(g[a[i][p]-'a'][a[i+1][p]-'a']) continue;
    71         g[a[i][p]-'a'][a[i+1][p]-'a']=1;
    72         in[a[i+1][p]-'a']++;
    73     }
    74     topsort();
    75     if(jishu==26){
    76         for(int i=0;i<=25;i++)
    77             printf("%c",ans[i]+'a');
    78     }
    79     else
    80         printf("Impossible
    ");
    81     return 0;
    82 }
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  • 原文地址:https://www.cnblogs.com/hsd-/p/7295450.html
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