ACM/ICPC 之 欧拉回路两道(POJ1300-POJ1386)

两道有关欧拉回路的例题

---

POJ1300-Door Man

//判定是否存在从某点到0点的欧拉回路
//Time:0Ms	Memory:116K
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

#define MAX 25
int st, n;
int door[MAX];

int main()
{
	char s[120];
	while (scanf("%s", s), strcmp(s, "ENDOFINPUT"))
	{
		memset(door, 0, sizeof(door));
		int doors = 0;
		scanf("%d%d", &st, &n);
		gets_s(s, 120);
		for (int i = 0; i < n; i++)
		{
			gets_s(s, 120);
			int num, k = 0;
			while (sscanf(s + k, "%d", &num) == 1)
			{
				doors++;
				door[num]++;
				door[i]++;
				while (s[k] == ' ')	k++;
				while (s[k] && s[k] != ' ') k++;
			}
		}
		gets_s(s, 120);
		int odd = 0;
		for (int i = 0; i < n; i++)
			odd += door[i] % 2 == 1;
		if (odd == 0 && st == 0)	
			printf("YES %d
", doors);	//无奇度节点
		else if (odd == 2 && st && door[st] % 2 && door[0] % 2)
			printf("YES %d
", doors);	//两个奇度节点
		else printf("NO
");
	}
	return 0;
}

---

POJ1386-Plays on Words

//判断能否使给定的词组前后接龙
//Time:344Ms	Memory:120K
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

#define MAX 28
#define MAXS 1005
#define MAXN 100005

int n;
char s[MAXS];
int in[MAX], out[MAX];
int fa[MAX];

int find(int x)
{
	return fa[x] < 0 ? x : find(fa[x]);
}

int Union(int r1, int r2)
{
	r1 = find(r1);	r2 = find(r2);
	if (r1 == r2)	return r1;
	int tmp = fa[r1] + fa[r2];
	if (fa[r1] > fa[r2])
	{
		fa[r1] = r2;
		fa[r2] = tmp;
		return r2;
	}
	else {
		fa[r2] = r1;
		fa[r1] = tmp;
		return r1;
	}
}

int main()
{
	//freopen("words.in", "r", stdin);
	int T;
	scanf("%d", &T);
	while (T--) {
		memset(in, 0, sizeof(in));
		memset(out, 0, sizeof(out));
		memset(fa, -1, sizeof(fa));

		int pa;
		scanf("%d", &n);
		while (n--) {
			scanf("%s", s);
			int i = s[strlen(s) - 1] - 'a';
			int o = s[0] - 'a';
			out[o]++;	in[i]++;
			pa = Union(i, o);
		}

		int odd = 0;
		bool connect = true;
		bool A = false, B = false;
		for (int i = 0; i < 26; i++)
		{
			if (!in[i] && !out[i]) continue;
			if (pa != find(i)) {
				connect = false;	break;
			}
			if (in[i] - out[i] != 0)
			{
				odd++;
				if (in[i] - out[i] == 1)	A = true;
				if (in[i] - out[i] == -1)	B = true;
			}
		}
		if (connect && ((odd == 2 && A && B) || odd == 0))
			printf("Ordering is possible.
");
		else printf("The door cannot be opened.
");
	}

	return 0;
}