拓扑排序模板（题）

拓扑排序

 

拓扑图：有向⽆环图

在拓扑序中，所有指向x点的边的起点，都在x点之前。

 

伪代码：

将⼊度为0的点加⼊队列 

while(l<r){ 

　　 int x=q[++l]; 

　　 for(int i=point[x];i;i=next[i]) //这是邻接表的写法，也可以转化为vector

　　{

　　　　in[to[i]]- -;

　　　　if(in[to[i])==0)

　　　　q[++r]=to[i]; 

　　 }

 

/*

CF510C Fox And Names
题意翻译

记得熟悉的字典序吗？ 现在有一些按照某个被改变的字典序（26个字母的顺序不再是abcdefg...xyz）排序后的字符串，请求出其字典序。
（存在多组解，请输出任意一组）
否则输出 Impossible（请注意大小写）

数据范围 字符串很开心自己有至多100个字符

其他限制 见右边
题目描述

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 s s and t t t , first we find the leftmost position with differing characters: si≠ti s_{i}≠t_{i} si?≠ti? . If there is no such position (i. e. s s s is a prefix of t t t or vice versa) the shortest string is less. Otherwise, we compare characters si s_{i} si? and ti t_{i} ti? according to their order in alphabet.
输入输出格式
输入格式：

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

Each of the following n n n lines contain one string namei name_{i} namei? ( 1<=∣namei∣<=100 1<=|name_{i}|<=100 1<=∣namei?∣<=100 ), the i i i -th name. Each name contains only lowercase Latin letters. All names are different.

输出格式：

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).

输入输出样例
输入样例#1： 复制

3
rivest
shamir
adleman

输出样例#1： 复制

bcdefghijklmnopqrsatuvwxyz

输入样例#2： 复制

10
tourist
petr
wjmzbmr
yeputons
vepifanov
scottwu
oooooooooooooooo
subscriber
rowdark
tankengineer

输出样例#2： 复制

Impossible

输入样例#3： 复制

10
petr
egor
endagorion
feferivan
ilovetanyaromanova
kostka
dmitriyh
maratsnowbear
bredorjaguarturnik
cgyforever

输出样例#3： 复制

aghjlnopefikdmbcqrstuvwxyz

输入样例#4： 复制

7
car
care
careful
carefully
becarefuldontforgetsomething
otherwiseyouwillbehacked
goodluck

输出样例#4： 复制

acbdefhijklmnogpqrstuvwxyz



*/
#include<bits/stdc++.h>
using namespace std;
struct edge{
    int next,to;
}e[1005];
int point[1005],cnt;
int l,r,q[1005],n;
char s[110][110];
int in[1005];
void addedge(int x,int y){
    e[++cnt].next=point[x];
    e[cnt].to=y;
    point[x]=cnt;
}
void toposort(){
    l=r=0;
    for(int i=0;i<26;i++)
        if(in[i]==0)
            q[++r]=i;
    while(l<r){
        int x=q[++l];
        for(int i=point[x];i;i=e[i].next){//用vector就直接<size()就行了，不用next 
            int y=e[i].to;
            in[y]--;
            if(in[y]==0)
                q[++r]=y;
        }
    }//队列里就是拓扑序 
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%s",s[i]);
    for(int i=1;i<n;i++){
        int len1=strlen(s[i]);
        int len2=strlen(s[i+1]);
        bool flag=0;
        for(int j=0;j<min(len1,len2);j++)
            if(s[i][j]==s[i+1][j])
                continue;
            else{
                flag=1;
                in[s[i+1][j]-'a']++;
                addedge(s[i][j]-'a',s[i+1][j]-'a');
                break;
            }
        if(flag==0 &&len2<len1){
            printf("Impossible");
            return 0;
        }
    }
    toposort();
    if(l<26)
        printf("Impossible");
    else
        for(int i=1;i<=26;i++)
            printf("%c",q[i]+'a');

        
    
    return 0;
 }