Codeforces 510C

Posted 2020-10-01 %%%%%

CodeForces 510C

题意：给出n个名字，问是否存在一个字典序（这个字典序是任意的）使得这些名字是按这个字典序输出的，如果存在输出任意一个可行的字典序

思路：拓扑排序，关键是建图，第i个和i+1个建一条由i指向i+1的单向边，建图是a-z26个字母间建图，每2个相邻的单词之间排序的依据是第一个不相同的字母，由这个不相同的字母之间建边，如果没有不相同的则不建边，这里有一个坑，就是 aaa aaaa这种数据，要特判

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define ll long long
#define endl ("\n")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ft (frist)
#define sd (second)
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

struct AN{
    int id,num;
    bool friend operator< (AN a, AN b){
        return a.num<b.num;
    }
}ans[160];
char s[155][155];
int n,head[155],nex[305],to[305],tot=1,in[300],cnt;
bool vis[155];
void add(int u, int v){
    to[tot]=v;
    nex[tot]=head[u];
    head[u]=tot++;
}

void topu(int u){
    queue<int> Q;
    while(!Q.empty()) Q.pop();
    Q.push(u);
    vis[u]=1;
    while(!Q.empty()){
        int now=Q.front(); Q.pop();
        ans[now].num=++cnt;
        for(int i=head[now]; i!=-1; i=nex[i]){
            int v=to[i]; in[v]--;
            if(in[v]==0 && vis[v]==0){
                Q.push(v);
                vis[v]=1;
            }
        }
    }
}
int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n;
    mem(head,-1);
    for(int _=1; _<=n; ++_){
        cin>>s[_];
    }
    for(int i=1; i<n; ++i){
        int k=0;
        int l1=strlen(s[i]), l2=strlen(s[i+1]);
        while(s[i][k]==s[i+1][k] && k<=l1 && k<=l2){
            ++k;
        }
        if(k>=l1 || k>=l2){
            if(l2<l1){
                cout<<"Impossible\n";
                return 0;
            }
            continue;
        }
        add(s[i][k]-‘a‘+1, s[i+1][k]-‘a‘+1);
        in[s[i+1][k]-‘a‘+1]++;
    }
    for(int i=1; i<=26; ++i){ //cout<<in[i]<<" ";
        if(in[i]==0 && vis[i]==0) topu(i);//bug(ans[i].num)
    }
    for(int i=1; i<=26; ++i){
        if(ans[i].num==0){
            cout<<"Impossible\n";
            return 0;
        }
    }
    for(int i=1; i<=26; ++i) ans[i].id=i;
    sort(ans+1,ans+1+26);
    for(int i=1; i<=26; ++i) cout<<char(ans[i].id+‘a‘-1);
    return 0;
}