poj 2337 Catenyms 【欧拉路径】

题目链接：http://poj.org/problem?id=2337

题意：给定一些单词，假设一个单词的尾字母与还有一个的首字母同样则能够连接。问能否够每一个单词用一次，将全部单词连接，能够则输出字典序最小的序列。

代码：
（bin 神的板子）

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

using namespace std;

int t;
int n;
string s[1010];

struct Edge
{
    int to, next;
    int index;
    bool flag;
}edge[2010];

int head[300], 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++;
}

int in[250], out[250];

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;
    cin >> t;
    while (t--)
    {

        scanf("%d",&n);
        for (int i = 0; i < n; i++)
            cin >> s[i];
        sort(s, s + n);
        init();
        memset(in,0,sizeof(in));
        memset(out, 0, sizeof(out));
        int start = 100;
        for (int i = n - 1; i >= 0; i--)
        {
            int u = s[i][0] - ‘a‘;
            int v = s[i][s[i].length() - 1] - ‘a‘;
            addedge(u,v,i);
            out[u]++;
            in[v]++;
            if (u < start) start = u;
            if (v < start) start = v;
        }
        int cc1 = 0, cc2 = 0;
        for (int i = 0; i < 26; i++)
        {
            if (out[i] - in[i] == 1)
            {
                cc1++;
                start = i;
            }
            else if (out[i] - in[i] == -1)
                cc2++;
            else if (out[i] - in[i] != 0)
                cc1 = 3;
        }
        if (!((cc1 == 0 && cc2 == 0) || (cc1 == 1 && cc2 == 1)))
        {
            printf("***\n");
            continue;
        }
        cnt = 0;
        dfs(start);
        if (cnt != n)
        {
            printf("***\n");
            continue;
        }
        for (int i = n-1; i >=0 ; i--)
        {
            cout << s[ans[i]];
            if (i != 0) printf(".");
            else printf("\n");
        }
    }
    return 0;
}