呼~~终于把它A掉了
这道题是判断是否存在欧拉回路的，把颜色看成点，珍珠看成边，然后统计每一个点的度，如果有奇度就表明欧拉回路不存在，如果存在的话直接dfs出欧拉回路，然后扔到栈里最后直接输出就行了。（仰慕佳哥）
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <iostream>
using namespace std;
const int N = 55;
int g[N][N], dg[N];
stack<pair<int, int> >st;
int is_euler(int mn){
int s = -1;
for (int i = 1; i <= mn; i++){
if (dg[i]){
if (s == -1) s = i;
if (dg[i] % 2) return -1;
}
}
return s;
}
void dfs(int cur, int mn){
for (int i = 1; i <= mn; i++){
if (g[cur][i]){
g[cur][i] -= 1; g[i][cur] -= 1;
dfs(i, mn);
st.push(make_pair(cur, i));
}
}
}
int main(){
//freopen("out.txt", "w", stdout);
int cs, n;
scanf("%d", &cs);
for (int p = 1; p <= cs; p++){
memset(g, 0, sizeof(g));
memset(dg, 0, sizeof(dg));
if (p > 1) printf("\n");
scanf("%d", &n);
int x, y, mn = 0;
for (int i = 0; i < n; i++){
scanf("%d%d", &x, &y);
g[x][y] += 1; g[y][x] += 1;
dg[x] += 1; dg[y] += 1;
mn = max(mn, max(x, y));
}
int f = is_euler(mn);
printf("Case #%d\n", p);
if (f == -1){
puts("some beads may be lost");
}
else{
dfs(f, mn);
while(!st.empty()){
printf("%d %d\n", st.top().first, st.top().second);
st.pop();
}
}
}
return 0;
}