UVA 10054 The Necklace

UVA 10054 The Necklace

依然是欧拉回路。不过要多一部输出顺序，用递归输出即可。

判断是否都连通并且是否每个点度数都为偶数

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

int t;
int tt;
int n;
int parent[55];
int du[55];
int vis[55];
int judge;
int v[55][55];
int chu[55];
int ru[55];
int find(int x)
{
    if (x != parent[x])
	return find(parent[x]);
    else
	return x;
}

void uion(int a, int b)
{
    int pa = find(a);
    int pb = find(b);
    if (pa != pb)
	parent[pa] = pb;
}

void dfs(int x)
{
    for (int i = 1; i <= 50;  i++)
    {
	if(v[x][i])
	{
	    v[x][i] --;
	    v[i][x] --;
	    dfs(i);
	    printf("%d %d\n", i, x);
	}
    }
}
int main()
{
    scanf("%d", &t);
    for (tt = 1; tt <= t; tt ++)
    {
	int a, b;
	int max = 0;
	int star;
	judge = 0;
	memset(du, 0, sizeof(du));
	memset(parent, 0, sizeof(parent));
	memset(vis, 0, sizeof(vis));
	memset(v, 0, sizeof(v));
	memset(chu, 0, sizeof(chu));
	memset(ru, 0, sizeof(ru));
	scanf("%d", &n);
	for (int i = 1; i <= 50; i ++)
	    parent[i] = i;
	for (int i = 0; i < n; i ++)
	{
	    scanf ("%d%d", &a, &b);
	    v[a][b] ++;
	    v[b][a] ++;
	    chu[b] ++;
	    ru[a] ++;
	    //du[a] ++;
	    //du[b] ++;
	    vis[a] = vis[b] = 1;
	    uion(a, b);
	}
	for (int i = 1; i <= 50; i ++)
	{
	    //if (du[i] % 2)
	    if ((chu[i] + ru[i]) % 2)
	    {
		judge = 1;
		break;
	    }	
	    //else if(du[i] > max)
	    else if ((chu[i] + ru[i]) > max)
	    {
		//max = du[i];
		max = chu[i] + ru[i];
		star = i;
	    }
	}
	if (judge == 0)
	{
	    for (int i = 1; i <= 50; i ++)
	    {
		if (vis[i])
		{
		    for (int j = 1; j <= 50; j ++)
		    {
			if (vis[j])
			{
			    if (find(i) != find(j))
			    {
				judge = 1;
				break;
			    }	
			}
		    }
		}
		if (judge)
		    break;
	    }
	}
	printf("Case #%d\n", tt);
	if (judge)
	{
	    printf("some beads may be lost\n");
	}
	else
	{
	    dfs(a);
	}
	if (tt != t)
	    printf("\n");

    }
    return 0;
}