首页 > 其他 > 详细

POJ 2230 Watchcow

时间:2015-08-09 22:05:56      阅读:145      评论:0      收藏:0      [点我收藏+]

Watchcow

Time Limit: 3000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 2230
64-bit integer IO format: %lld      Java class name: Main
 
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

Hint

OUTPUT DETAILS: 

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

Source

 
解题:欧拉回路
 
技术分享
 1 #include <cstdio>
 2 #include <cstring>
 3 using namespace std;
 4 const int maxn = 10010;
 5 struct arc {
 6     int to,next;
 7     arc(int x = 0,int y = -1) {
 8         to = x;
 9         next = y;
10     }
11 } e[500000];
12 int head[maxn],tot,n,m;
13 bool vis[500000];
14 void add(int u,int v) {
15     e[tot] = arc(v,head[u]);
16     head[u] = tot++;
17 }
18 void dfs(int u) {
19     for(int &i = head[u]; ~i; i = e[i].next) {
20         if(vis[i]) continue;
21         vis[i] = true;
22         dfs(e[i].to);
23     }
24     printf("%d\n",u);
25 }
26 int main() {
27     int u,v;
28     while(~scanf("%d%d",&n,&m)) {
29         memset(head,-1,sizeof head);
30         memset(vis,false,sizeof vis);
31         for(int i = tot = 0; i < m; ++i) {
32             scanf("%d%d",&u,&v);
33             add(u,v);
34             add(v,u);
35         }
36         dfs(1);
37     }
38     return 0;
39 }
View Code

 

POJ 2230 Watchcow

原文:http://www.cnblogs.com/crackpotisback/p/4716400.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!