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NC51343. Watchcow

描述

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 bidirectional trails numbered between 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.

输入描述

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

输出描述

* 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.

示例1

输入:

4 5
1 2
1 4
2 3
2 4
3 4

输出:

1
2
3
4
2
1
4
3
2
4
1

说明:

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

原站题解

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C++14(g++5.4) 解法, 执行用时: 32ms, 内存消耗: 3236K, 提交时间: 2019-12-06 11:13:27

#include<bits/stdc++.h>
using namespace std;
char buf[1<<20],*_=buf,*__=buf;
#define gc() (_==__&&(__=(_=buf)+fread(buf,1,1<<20,stdin),_==__)?EOF:*_++)
#define TT template<class T>inline
TT bool read(T &x){
    x=0;char c=gc();
    while(c<48||c>57)c=gc();
    while(47<c&&c<58)x=(x<<3)+(x<<1)+(c^48),c=gc();
    return 1;
}
TT bool read(T&a,T&b){return read(a)&&read(b);}
const int MAXN=1e4+8;
struct E{int y,nt;}e[MAXN*10];
int head[MAXN],cnt;
inline void add(int x,int y){
    e[++cnt].y=y;
    e[cnt].nt=head[x];
    head[x]=cnt;
}
int n,m;
void dfs(int x){
    for(int i=head[x];head[x]=e[i].nt,i;i=head[x]){
        head[x]=e[i].nt;
        dfs(e[i].y);
    }
    printf("%d\n",x);
}
int main() {
    read(n,m);
    for(int i=0,x,y;i<m;++i)read(x,y),add(x,y),add(y,x);
    dfs(1);
    return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 23ms, 内存消耗: 1656K, 提交时间: 2020-06-08 14:23:16

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;
int head[maxn],Next[maxn],to[maxn],tot;
int st[maxn],ans[maxn];bool vis[maxn];
int n,m,top,t;
void add(int x,int y)
{
	to[++tot]=y,Next[tot]=head[x],head[x]=tot;
}
void euler()
{
	st[++top]=1;
	while(top>0){
		int x=st[top],i=head[x];
		//while(i&&vis[i])	i=Next[i];
		if(i){
			st[++top]=to[i];
			//vis[i]=vis[i^1]=true;
			head[x]=Next[i];
		}else{
			top--;ans[++t]=x;
		}
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int t1,t2;scanf("%d%d",&t1,&t2);
		add(t1,t2);add(t2,t1);
	}
	euler();
	for(int i=t;i;i--){
		printf("%d\n",ans[i]);
	}
}

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