UVA10054 UVALive2036 The Necklace【欧拉回路】

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 recolle ct 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 (5 ≤ N ≤ 1000) 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 1 ≤ i ≤ N − 1, 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

Regionals 2000 >> Asia - Shanghai

问题链接：UVA10054 UVALive2036 The Necklace
问题简述：（略）
问题分析：
    欧拉回路问题，输出欧拉回路。
    先统计出入度，然后进行欧拉回路的判定和处理。存在欧拉回路时则输出。回路输出用DFS实现。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10054 UVALive2036 The Necklace */

#include <bits/stdc++.h>

using namespace std;

const int N = 50;
const int M = 1000;
int degree[M + 1], g[M+ 1][M + 1];

void print(int u)
{
    for(int v = 1; v <= N; v++)
        if(g[u][v] > 0) {
            g[u][v]--;
            g[v][u]--;
            print(v);
            printf("%d %d\n", v, u);
        }
}

void output()
{
    for(int i = 1; i <= N; i++)
        if(degree[i]) {
            print(i);
            break;
        }
}

int main()
{
    int t, caseno = 0, n;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);

        memset(g, 0, sizeof(g));
        memset(degree, 0, sizeof(degree));

        // 读入边，统计各个结点间边数量，统计各个结点出入度（无向图）
        for(int i = 1; i <= n; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            g[u][v]++;
            g[v][u]++;
            degree[u]++;
            degree[v]++;
        }

        bool flag = false;
        for(int i = 1; i <= N; i++)
            if(degree[i] % 2) {
                flag = true;
                break;
            }

        printf("Case #%d\n", ++caseno);
        if(flag)
            printf("some beads may be lost\n");
        else
            output();
        printf("\n");
    }

    return 0;
}