poj

题解思路：因为要字典序最小所以要先排序一下，然后就套欧拉。

代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>using namespace std;struct Edge{    int to,next;    int index;    bool flag;}edge[2010];int head[30],tot;void init(){    tot = 0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int index){    edge[tot].to = v;    edge[tot].next = head[u];    edge[tot].index = index;    edge[tot].flag = false;    head[u] = tot++;}string str[1010];int in[30],out[30];int cnt;int ans[1010];void dfs(int u){    for(int i = head[u] ;i != -1;i = edge[i].next)        if(!edge[i].flag)        {            edge[i].flag = true;            dfs(edge[i].to);            ans[cnt++] = edge[i].index;        }}int main(){    int T,n;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        for(int i = 0;i < n;i++)            cin>>str[i];        sort(str,str+n);//要输出字典序最小的解，先按照字典序排序        init();        memset(in,0,sizeof(in));        memset(out,0,sizeof(out));        int start = 100;        for(int i = n-1;i >= 0;i--)//字典序大的先加入        {            int u = str[i][0] - 'a';            int v = str[i][str[i].length() - 1] - 'a';            addedge(u,v,i);            out[u]++;            in[v]++;            start = min(start,min(u,v));        }        int cc1 = 0, cc2 = 0;        for(int i = 0;i < 26;i++){            if(out[i] - in[i] == 1){                cc1++;                start = i;//如果有一个出度比入度大1的点，就从这个点出发，否则从最小的点出发            }            else if(out[i] - in[i] == -1)                cc2++;            else if(out[i] - in[i] != 0)                cc1 = 2;        }        if(!((cc1 == 0 && cc2 == 0)||(cc1 == 1 && cc2 == 1)))        {            printf("***\n");            continue;        }        cnt = 0;        dfs(start);        if(cnt != n)//判断是否连通        {            printf("***\n");            continue;        }        for(int i = cnt-1; i >= 0;i--)        {            cout<<str[ans[i]];            if(i > 0)printf(".");            else printf("\n");        }    }    return 0;}