UVa 10054 - The Necklace (欧拉回路)

题目链接：​​https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=995&mosmsg=Submission+received+with+ID+26578658​​

如果把每个珠子看成点，不好建图。

将每种颜色看成点，每个珠子就对应一条无向边，于是转化成求欧拉回路的问题。

无向图求欧拉回路要注意经过一条边以后，将反向边也标记。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 5010;
const int maxcolor = 50;

int T, n;

int h[maxcolor+10], cnt = 1;
struct e{
  int from, to, id, next;
}e[maxn]; 
void add(int u, int v, int id){
  e[++cnt].to = v;
  e[cnt].from = u;
  e[cnt].id = id; // 项链的编号
  e[cnt].next = h[u];
  h[u] = cnt; 
}

int vis[maxn], deg[maxcolor+10];
vector<int> path;

void euler(int u){
  for(int i = h[u] ; i != -1 ; i = e[i].next){
    if(!vis[i]){
      vis[i] = 1;
      vis[i ^ 1] = 1; 
      int v = e[i].to;
      euler(v);
      path.push_back(i);
    }
  }
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
  int kase = 0;
  int flag = 0; 
  T = read();
  while(T--){
    if(flag) printf("\n");
    flag = 1;
    path.clear();
    memset(deg, 0, sizeof(deg));
    memset(h, -1, sizeof(h)); cnt = 1;
    memset(vis, 0, sizeof(vis));
    
    n = read();
    
    int u, v;
    int start = -1;
    for(int i = 1 ; i <= n ; ++i){
      u = read(), v = read();
      ++deg[u]; ++deg[v];
      add(u, v, i); add(v, u, i);
      start = u;
    }
    bool solved = true;
    
    for(int i = 1 ; i <= maxcolor ; ++i){
      if(deg[i] % 2 == 1){
        solved = false;
        break;
      }
    } 
    
    if(solved) euler(start);
    
    printf("Case #%d\n", ++kase);
    if(path.size() != n ||e[path[0]].to != e[path[path.size()-1]].from){
      solved = false;
    } 
    if(!solved) printf("some beads may be lost\n");
    else {
      for(int i = path.size() - 1 ; i >= 0 ; --i){
        printf("%d %d\n", e[path[i]].from, e[path[i]].to);
      }     
    }
  }
  return 0;
}