uva10129 play on words

#include <stdio.h>
#include <string.h>
#include <math.h>
char s[1005];
int in[27], out[27];
int g[27][27];
int vis[27];
//int same[27];

int Euler()//返回1表示在图全连通的条件下，是欧拉道路
{
    int n = 0;
    for (int i = 1;i <= 26;i++)
    {
        //if (same[i] && !in[i] && !out[i])return 0;
        int sum = in[i] + out[i];
        if (sum == 1 || sum == -1)n++;
        else if (sum != 0)return 0;
    }
    if (n != 0 && n != 2)return 0;
    return 1;
}

void dfs(int u)
{
    for (int i = 1;i <= 26;i++)
    {
        if (!vis[i] && g[i][u])
        {
            vis[i] = 1;
            dfs(i);
        }
    }
}

int DFS()//返回1表示图全连通
{
    int count = 0;
    for (int i = 1;i <= 26;i++)
    {
        if (in[i] || out[i])
            if (!vis[i])
            {
                vis[i] = 1;
                if (++count > 1)return 0;
                dfs(i);
            }
    }
    return 1;
}

int main(void)
{
    int t, n;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        memset(in, 0, sizeof(in));
        memset(out, 0, sizeof(out));
        // memset(same, 0, sizeof(same));
        memset(vis, 0, sizeof(vis));
        memset(g, 0, sizeof(g));
        while (n--)
        {
            scanf("%s", s);
            int _last = strlen(s) - 1;
            int u = s[0] - 'a' + 1, v = s[_last] - 'a' + 1;
            //uv相同不需特别处理，否则当全部是比如mem这种就会出错,难过，自以为谨小慎微反而错了--bug了好久好久
            g[u][v] = g[v][u] = 1;
            // if (u == v)same[u] = 1;
            //else {
            in[u]++;
            out[v]--;
            //}
        }
        int a = !DFS(), b = !Euler();
        if (a || b)
            printf("The door cannot be opened.\n");
        else
            printf("Ordering is possible.\n");
    }
    return 0;
}