POJ 2330 Watchcow（欧拉回路）

最新推荐文章于 2019-07-31 11:03:51 发布

原创最新推荐文章于 2019-07-31 11:03:51 发布 · 350 阅读

0 ·

CC 4.0 BY-SA版权

暑期集训同时被 2 个专栏收录

18 篇文章

订阅专栏

图论

6 篇文章

订阅专栏

本文介绍了一种求解欧拉回路的问题背景与算法实现方法。在一个有向图中，如何找到一条能够遍历每条边恰好一次并回到起点的路径。通过使用深度优先搜索（DFS）和栈结构，文章提供了一个有效的解决方案。

Watchcow

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8249		Accepted: 3589		Special Judge

Description

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

Sample Output

Hint

OUTPUT DETAILS:

Bessie starts at 1 (barn), goes to 2, then 3, etc...

Source

USACO 2005 January Silver

题意：给出一有向图，输出其欧拉回路

#include <stdio.h>
#include <vector>
#include <stack>
using namespace std;

int n, m;
struct xx{
    int v, vis;
};
vector<xx> k[10010];
stack<int> s;

void dfs(int x){
    for(int i = 0; i < k[x].size(); i++){
        if(!k[x][i].vis){
            k[x][i].vis = 1;
            dfs(k[x][i].v);
            s.push(k[x][i].v);
        }
    }
}

int main(){
    while(scanf("%d%d", &n, &m) == 2){
        for(int i = 0; i < m; i++){
            int x, y;
            xx s;
            scanf("%d%d",&x ,&y);
            s.v = y, s.vis = 0;
            k[x].push_back(s);
            s.v = x;
            k[y].push_back(s);
        }
        dfs(1);
        s.push(1);
        while(!s.empty()){
            printf("%d\n", s.top());
            s.pop();
        }
    }
}