uva-10129 單詞

成語接龍可以抽象爲歐拉回路，單詞爲有向邊，詞頭詞尾爲節點，用dfs判斷是否圖是否連通，先判斷入度出度，然後dfs，在判斷圖是否連通。visitedv數組用來區別不存在的點、存在的點和存在且已用dfs訪問過的點

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <iostream>
using namespace std;
#define DEBUG
const int maxn=1000+5,maxv=26;
int n,startcnt,endcnt,startv,endv;
char s[maxn];
int in[26],out[26],visitedv[26];
bool visited[26][26],G[26][26],flag;
bool dfs(int u){
    visitedv[u]=2;
    for(int v=0;v<maxv;v++){
        if(G[u][v]&&!visited[u][v]){
            visited[u][v]=1;
            dfs(v);
            // printf("%d %d\n",u,v);
        }
    }
    return true;
}
int main(){
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int m;
    cin>>m;
    while(m--){
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(visited,0,sizeof(visited));
        memset(G,0,sizeof(G));
        memset(visitedv,0,sizeof(visitedv));
        startcnt=0;
        endcnt=0;
        cin>>n;
        while(n--){
            cin>>s;
            int len=strlen(s);
            visitedv[s[0]-'a']=visitedv[s[len]-'a']=1;
            G[s[0]-'a'][s[len-1]-'a']=1;
            in[s[0]-'a']++;out[s[len-1]-'a']++;
        }
        flag=1;
        for(int i=0;i<26;i++){
            if(in[i]!=out[i]){
                if(in[i]==out[i]+1)
                    startv=i,startcnt++;
                else if(in[i]+1==out[i])
                    endv=i,endcnt++;
                else{flag=0;break;}
            }
            if(!startcnt&&in[i])
                startv=i;
        }
        if(endcnt>1||startcnt>1)flag=0;
        if(!flag){cout<<"The door cannot be opened.\n";continue;}
        // printf("startv:%d\n",startv);
        // printf("endv:%d\n",endv);
        dfs(startv);
        flag=1;
        for(int i=0;i<26;i++)
            if(visitedv[i]==1){flag=0;cout<<"The door cannot be opened.\n";break;}
        if(flag)
        cout<<"Ordering is possible.\n";

    }
#ifdef DEBUG
    fclose(stdin);
    fclose(stdout);
#endif
    return 0;
}