UVA The Necklace （并查集判断欧拉回路）

Problem D: The Necklace

My little sister had a beautiful necklace made of colorful beads. Two successive beads in the necklace shared a common color at their meeting point. The figure below shows a segment of the necklace:



But, alas! One day, the necklace was torn and the beads were all scattered over the floor. My sister did her best to recollect all the beads from the floor, but she is not sure whether she was able to collect all of them. Now, she has come to me for help. She
wants to know whether it is possible to make a necklace using all the beads she has in the same way her original necklace was made and if so in which order the bids must be put.

Please help me write a program to solve the problem.

Input

The input contains T test cases. The first line of the input contains the integer T.

The first line of each test case contains an integer N (

) giving the number of beads my sister was able to collect.
Each of the next N lines contains two integers describing the colors of a bead. Colors are represented by integers ranging from 1 to 50.

Output

For each test case in the input first output the test case number as shown in the sample output. Then if you apprehend that some beads may be lost just print the sentence ``some beads may be lost" on a line by itself. Otherwise, print N lines
with a single bead description on each line. Each bead description consists of two integers giving the colors of its two ends. For

,
the second integer on line i must be the same as the first integer on line i + 1. Additionally, the second integer on line N must be equal to the first integer on line 1. Since there are many solutions, any one of them is acceptable.

Print a blank line between two successive test cases.

Sample Input

2
5
1 2
2 3
3 4
4 5
5 6
5
2 1
2 2
3 4
3 1
2 4

Sample Output

Case #1
some beads may be lost
 
Case #2
2 1
1 3
3 4
4 2
2 2

Miguel Revilla

2000-12-28

题意：有一个项链其中的每一个珠子都有两种颜色，每个珠子相连的规则是其中的珠子的两种颜色都能匹配到和他颜色一样的珠子，如题目所给的图所示。现在这样的一串项链的珠子散落了，只找到了其中的一部分，问是否可以用找到的这一部分珠子重新做一个和原来项链规则一样的。能的话输出新项链的组成顺序，（输出不唯一），不能输出some .....；

并查集判断欧拉回路。

代码：

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

const int M = 100;

int num[M];
int map[M][M];
int n;

int findx(int x)
{
    int r = x;
    while(r!=num[r])
    {
        r = num[r];
    }
    int k = x,j;
    while(k!=r)
    {
        j = num[k];
        num[k] = r;
        k = j;
    }
    return r;
}

void bing(int x,int y)
{
    int fx = findx(x);
    int fy = findx(y);
    if(fx!=fy)
    {
        num[fx] = fy;
    }

}

void print(int k)
{
    for (int i = 0; i < M; i++)
    {
        if (map[k][i])
        {
            map[k][i]--;
            map[i][k]--;
            print(i);
            printf("%d %d\n", i,k);
        }
    }
}

int main()
{
    int t, k = 1;
    int f[M];
    scanf("%d" ,&t);
    while (t--)
    {
        memset(f, 0, sizeof(f));
        memset(map, 0, sizeof(map));
        int sum = 0;
        for (int i = 0; i < M; i++)
        {
            num[i] = i;
        }

        scanf("%d", &n);
        for (int i = 0; i < n; i++)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            f[a]++;
            f[b]++;
            map[a][b]++;
            map[b][a]++;
            bing(a,b);
        }

        int begin = 0;
        for (int i = 0; i < M; i++)
        {
            if (f[i] && findx(i) == i)
            {
                begin = i;
                break;
            }
        }

        for (int i = 0; i < M; i++)
        {
            //printf("num[%d] = %d\n",i,num[i]);
            sum += f[i] % 2;
            if (f[i] && begin != findx(i))
            {
                sum++;
            }
        }
        printf("Case #%d\n", k++);
        if (sum > 0)
        {
            printf("some beads may be lost\n");
        }
        else
        {
           // printf("begin = %d\n",begin);
            print(begin);
        }
        if (t!=0)
        {
            printf("\n");
        }
    }
    return 0;
}