UVa 10129

题意

单词链接, 如acm, malform, mouse可以链接

思路

欧拉回路

1.图连通
2.图中所有点的出度==入度，或只有两个奇度点并且一个点的入度比出度大1（终点），另一个点的出度比入度大1（起点）

调试的时候遇到几个比较坑的数据 需要特殊处理判断一下

input
3
2
aa
aaa
1
ab
2
aa
bb
output
Ordering is possible.
Ordering is possible.
The door cannot be opened.

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int in[30], out[30], G[30][30], vis[30];

void dfs(int a)
{
    vis[a] = 1;
    for (int b = 0; b < 26; b++)
        if (!vis[b] && G[a][b])
            dfs(b);
}


int main()
{
    int T, n;
    char s[1000+10];
    scanf("%d",&T);
    while(T--){
        bool ok = true, ok2 = false;
        memset(vis, 0, sizeof(vis));
        memset(G, 0, sizeof(G));
        memset(in, 0, sizeof(in));
        memset(out, 0, sizeof(out));
        scanf("%d",&n);
        if(n==1)    ok2 = true;
        while(n--){
            scanf("%s",s);
            int len = strlen(s);
            int a = s[0]-'a', b = s[len-1]-'a';
            G[a][b]++;
            in[b]++, out[a]++;
        }
        if(ok2){
            puts("Ordering is possible.");
            continue;
        }
        int cnt1 = 0, cnt2 = 0, cnt = 0;
        for( int i = 0; i < 26; i++ ){
            if( in[i] == out[i] )   continue;
            if( in[i] + 1 == out[i] )   cnt1++;
            else if( out[i] + 1 == in[i] )  cnt2++;
            else{ ok = false;  break;}
        }
        if(cnt1 && cnt2 && cnt1+cnt2 > 2)   ok = false;
        int num = 0;
        if(ok){
            for( int i = 0; i < 26; i++ )
                if( out[i] ){
                    dfs(i);
                    break;
                }
            for( int i = 0; i < 26; i++ )
                if( in[i] + out[i] )
                    if( !vis[i] ){
                        ok = false;
                        break;
                    }
        }
        if(ok) puts("Ordering is possible.");
        else puts("The door cannot be opened.");
    }
    return 0;
}