uva10441 - Catenyms 欧拉通路

Problem C: Catenyms

A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the first letter of the second. For example, the following are catenyms:

dog.gopher
gopher.rat
rat.tiger
aloha.aloha
arachnid.dog

A compound catenym is a sequence of three or more words separated by periods such that each adjacent pair of words forms a catenym. For example,

aloha.aloha.arachnid.dog.gopher.rat.tiger

Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once. The first line of standard input contains t, the number of test cases. Each test case begins with 3 <= n <= 1000 - the number of words in the dictionary. n distinct dictionary words follow; each word is a string of between 1 and 20 lowercase letters on a line by itself. For each test case, output a line giving the lexicographically least compound catenym that contains each dictionary word exactly once. Output "***" if there is no solution.

Sample Input

2
6
aloha
arachnid
dog
gopher
rat
tiger
3
oak
maple
elm

Output for Sample Input

aloha.arachnid.dog.gopher.rat.tiger
***

  首尾字符相同的单词可以连起来，输出字典序最小的序列把所有串都连起来，如果不能连起来输出***。

  看着不难还做了好久。。

  欧拉通路：如果图中存在一条通过图中各边一次且仅一次的通路,则称此回路是欧拉通路。有向图欧拉通路判断条件是：图连通，一个点入度比出度大1，另一个点出度比入度大1，或者所有点出度等于入度。

  那么这里把字母当成点，单词当成边判断是否是欧拉通路就行，注意先判连通性。有一个问题是输出，我刚开始一直按顺序搜，遇到就输出，这样是不对的。比如za,ac,am,ma,cb，像我那么做直接是ba,ac,cb，而实际应该是za,am,ma,ac,cb。所以搜的时候如果找不到下一条边了就把这条边存起来，最后倒着输出，因为前面判过是欧拉通路了，最后肯定就是答案。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
#include<queue>
#include<map>
using namespace std;

typedef pair<int,int> pii;

const int MAXN=1010;
const int MAXM=25;
const int MAXNODE=32;
const int LOGMAXN=50;
const int INF=0x3f3f3f3f;

int T,N,K;
int vis[MAXN],ans[MAXN],in[30],out[30],alpha[30],top;
char a[MAXN][MAXM];
vector<int> G[30];
vector<pii> G2[30];

void dfs1(int u){
    vis[u]=1;
    int len=G[u].size();
    for(int i=0;i<len;i++){
        int v=G[u][i];
        if(!vis[v]) dfs1(v);
    }
}

void dfs2(int u){
    int len=G2[u].size();
    for(int i=0;i<len;i++){
        int e=G2[u][i].first;
        if(!vis[e]){
            vis[e]=1;
            dfs2(G2[u][i].second);
            ans[top++]=e;
        }
    }
}

int check(){
    int cnt=0,ret=0;
    for(int i=0;i<26;i++){
        if(in[i]==out[i]) continue;
        else if(out[i]-in[i]==1){
            cnt++;
            ret=i;
        }
        else if(in[i]-out[i]>1||out[i]-in[i]>1) return -1;
    }
    if(!cnt){
        for(int i=0;i<26;i++) if(vis[i]){
            ret=i;
            break;
        }
    }
    if(cnt<=1) return ret;
    return -1;
}

bool cmp(pii A,pii B){
    return strcmp(a[A.first],a[B.first])<0?true:false;
}

int main(){
    freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        K=0;
        memset(out,0,sizeof(out));
        memset(in,0,sizeof(in));
        memset(vis,0,sizeof(vis));
        memset(alpha,0,sizeof(alpha));
        for(int i=0;i<26;i++){
            G[i].clear();
            G2[i].clear();
        }
        int n,c;
        for(int i=1;i<=N;i++){
            scanf("%s",a[i]);
            int len=strlen(a[i]),u=a[i][0]-'a',v=a[i][len-1]-'a';
            out[u]++;
            in[v]++;
            if(!alpha[u]){
                K++;
                c=u;
                alpha[u]=1;
            }
            if(!alpha[v]){
                K++;
                c=v;
                alpha[v]=1;
            }
            G[u].push_back(v);
            G[v].push_back(u);
            G2[u].push_back(make_pair(i,v));
        }
        int cnt=0;
        dfs1(c);
        for(int i=0;i<26;i++) if(vis[i]){
            cnt++;
        }
        if(cnt==K) n=check();
        else n=-1;
        if(n==-1) printf("***\n");
        else{
            for(int i=0;i<26;i++) sort(G2[i].begin(),G2[i].end(),cmp);
            memset(vis,0,sizeof(vis));
            top=0;
            dfs2(n);
            for(int i=top-1;i>0;i--) printf("%s.",a[ans[i]]);
            printf("%s\n",a[ans[0]]);
        }
    }
    return 0;
}