博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
Til the Cows Come Home
阅读量:5282 次
发布时间:2019-06-14

本文共 2272 字,大约阅读时间需要 7 分钟。

Til the Cows Come Home

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 64   Accepted Submission(s) : 41
Problem Description
Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.
 
Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it. 
Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.
 

 

Input
* Line 1: Two integers: T and N
 
* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.
 

 

Output
* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.
 

 

Sample Input
5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100
 

 

Sample Output
90
 

 

Source
PKU
1 #include
2 #include
3 #define INF 999999 4 int Map[1006][1006]; 5 void SPFA(int s,int m,int *Dis) 6 { 7 int i,k,Start=0,End=1; 8 int Sign[10086],Visit[10086]; 9 Sign[Start] = s;10 Dis[s] = 0;11 while(Start
0&&Dis[i]>Map[k][i]+Dis[k])18 {19 Dis[i]=Dis[k]+Map[k][i];20 if(!Visit[i])21 {22 Sign[End++]=i;23 Visit[i]=1;24 }25 }26 }27 }28 }29 int main()30 {31 int i,T,N,a,b,c,Dis[10086];32 while(scanf("%d%d",&T,&N)!=EOF)33 {34 memset(Map,-1,sizeof(Map));35 memset(Dis,INF,sizeof(Dis));36 for(i=0;i
View Code

 

转载于:https://www.cnblogs.com/Wurq/articles/3929298.html

你可能感兴趣的文章
记叙在人生路上对你影响最大的三位老师
查看>>
002.大数据第二天
查看>>
python装饰器
查看>>
树上的路径
查看>>
【转载】TCP好文
查看>>
系统平均负载
查看>>
问题总结
查看>>
jenkins升级为2.134
查看>>
软件随笔
查看>>
C/C++知识补充 (1)
查看>>
Fast Poisson Disk Sampling
查看>>
Python Cookbook(第3版)中文版:15.14 传递Unicode字符串给C函数库
查看>>
Linux下SVN自动更新web [转]
查看>>
编程:对经验世界的析构与建构
查看>>
Openstack api 学习文档 & restclient使用文档
查看>>
vim linux下查找显示^M并且删除
查看>>
poj100纪念
查看>>
ExtJs4 笔记(5) Ext.Button 按钮
查看>>
把execl导入到数据库中
查看>>
阿里云人脸比对API封装
查看>>