USACO 3.3 fence 欧拉回路

题意：求给定图的欧拉回路（每条边只走一次）

 

若欧拉回路存在，图中只可能有0个or2个奇数度的点。

求解时，若有奇数度的点，则必须从该点开始。否则可以从任一点开始

求解过程：dfs

//主程序部分
# circuit is a global array
   find_euler_circuit
     circuitpos = 0
     find_circuit(node 1)
---------------------------------------------
# nextnode and visited is a local array
# the path will be found in reverse order
//递归函数
  find_circuit(node i)
    if node i has no neighbors then
      circuit(circuitpos) = node i
      circuitpos = circuitpos + 1
    else
      while (node i has neighbors)
          pick a random neighbor node j of node i
          delete_edges (node j, node i)
          find_circuit (node j)
      circuit(circuitpos) = node i
      circuitpos = circuitpos + 1
-------------------------------------------
最终结果：将circuit()数组倒序输出即可

 

 

/*
PROB:fence
LANG:C++
*/
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define INF 999999

int dx=INF,dy=0,n,f,x,y,p;
int e[1100][1100];
int t[1100],seq[1100];

void dfs(int x)
{
    for (int i=dx;i<=dy;i++)
        if (e[x][i]>0)
        {
            e[x][i]--;
            e[i][x]--;
            dfs(i);
        }
    p++;
    seq[p]=x;
}

int start()
{
    for (int i=dx;i<=dy;i++)
        if (t[i]%2!=0)
            return i;
    return 1;
}

int main()
{
    freopen("fence.in","r",stdin);
    freopen("fence.out","w",stdout);

    memset(e,0,sizeof(e));
    memset(t,0,sizeof(t));
    cin>>f;
    for (int i=1;i<=f;i++)
    {
        cin>>x>>y;
        e[x][y]++;
        e[y][x]++;
        t[x]++;
        t[y]++;
        if (x<dx)   dx=x;
        if (y<dx)   dx=y;
        if (x>dy)   dy=x;
        if (y>dy)   dy=y;

    }

    x=start();
    //cout<<dx<<" "<<dy<<"--"<<x<<endl;
    p=0;
    dfs(x);

    //cout<<p<<endl;
    for (int i=p;i>=1;i--)
        cout<<seq[i]<<endl;

    return 0;
}