Watchcow （poj-2230）

最新推荐文章于 2022-01-26 17:15:00 发布

原创最新推荐文章于 2022-01-26 17:15:00 发布 · 139 阅读

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本文介绍了一种解决特定欧拉回路问题的算法实现，通过使用邻接表存储图结构，实现从指定起点出发遍历所有边两次的路径寻找。代码示例清晰展示了深度优先搜索（DFS）的应用。

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Watchcow

Bessie’s been appointed the new watch-cow for the farm. Every night, it’s her job to walk across the farm and make sure that no evildoers are doing any evil. She begins at the barn, makes her patrol, and then returns to the barn when she’s done.

If she were a more observant cow, she might be able to just walk each of M (1 <= M <= 50,000) bidirectional trails numbered 1…M between N (2 <= N <= 10,000) fields numbered 1…N on the farm once and be confident that she’s seen everything she needs to see. But since she isn’t, she wants to make sure she walks down each trail exactly twice. It’s also important that her two trips along each trail be in opposite directions, so that she doesn’t miss the same thing twice.

A pair of fields might be connected by more than one trail. Find a path that Bessie can follow which will meet her requirements. Such a path is guaranteed to exist.

Input

Line 1: Two integers, N and M.
Lines 2…M+1: Two integers denoting a pair of fields connected by a path.

Output

Lines 1…2M+1: A list of fields she passes through, one per line, beginning and ending with the barn at field 1. If more than one solution is possible, output any solution.

Sample Input
4 5
1 2
1 4
2 3
2 4
3 4

Sample Output
1
2
3
4
2
1
4
3
2
4
1

首先该题说的很明显就是欧拉回路，而且还给你了起始点，我们还不需要通过入度找起点，只是数据比较大，那我们可以用邻接表存图，从1搜一下，最后在输出1就好了很简单

代码如下：

#include<stdio.h>
#include<string.h>
#include<stack>
#include<queue>
using namespace std;
const int maxn=10020;
//stack<int>f;
//queue<int>f;
int k=0,first[maxn],book[maxn];
struct node
{
    int x,next,flag;
} s[maxn*10];
void add(int x,int y)
{
    s[k].x=y,s[k].next=first[x];
    s[k].flag=0;
    first[x]=k++;
}
void dfs(int w)
{
   // f.push(w);
    for(int i=first[w]; i!=-1; i=s[i].next)
    {
        int x=s[i].x;
        if(s[i].flag==0)
        {
            s[i].flag=1;
            dfs(x);
        }
    }
    //f.push(w);
    printf("%d\n",w);
}
int main()
{
    int n,m,a,b;
    scanf("%d%d",&n,&m);
    memset(first,-1,sizeof(first));
    while(m--)
    {
        scanf("%d%d",&a,&b);
        add(a,b);  add(b,a);
    }
    dfs(1);
   /* while(!f.empty())
    {
        printf("%d\n",f.top());
        f.pop();
    }*/
    return 0;
}