uva 10129 Play on Words

uva 10129 Play on Words

欧拉回路，这题是有向图
判断方法是 点出入度都相等或者 有2个点出入度相差1，其他点都相等

思路：有100000个字符串，看似很大，其实可以只存字母的出入度，因为字母只有'a' - 'z'  26个， 只要开26的数组即可，然后判断度数是否符合条件

#include <stdio.h>
#include <string.h>

int t;
int n;
char sb[1005];
int parent[26];
int ru[26];
int chu[26];
int vis[26];
int judge;
int cent;
int find(int x)
{
if (x != parent[x])
return find(parent[x]);
else
return x;
}
void uion(int a, int b)
{
a = find(a);
b = find(b);
if (a != b)
parent[b] = a;
}
int main()
{
scanf("%d", &t);
while (t --)
{
cent = 0;
judge = 0;
for (int i = 0; i < 26; i ++)
parent[i] = i;
memset(ru, 0, sizeof(ru));
memset(chu, 0, sizeof(chu));
memset(vis, 0, sizeof(vis));
scanf("%d", &n);
getchar();
while (n --)
{
gets(sb);
int a = sb[strlen(sb) - 1] - 'a';
int b = sb[0] - 'a';
ru[b] ++;
chu[a] ++;
vis[a] = vis[b] = 1;
uion(a, b);
}
for (int i = 0; i < 26; i ++)
{
if (ru[i] != chu[i])
{
if (ru[i] - chu[i] == 1 || ru[i] - chu[i] == -1)
{
cent ++;
if (cent > 2)
{
judge = 1;
break;
}
}
else
{
judge = 1;
break;
}
}
}
for (int i = 0; i < 26; i ++)
{
if (vis[i])
{
for (int j = 0; j < 26; j ++)
{
if (vis[j])
{
if (find(i) != find(j))
{
judge = 1;
break;
}
}
}
}
if (judge)
break;
}
if (judge)
printf("The door cannot be opened.\n");
else
printf("Ordering is possible.\n");
}
return 0;
}