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



INPUT DETAILS:



There are five landmarks.


Sample Output



90



OUTPUT DETAILS:



Bessie can get home by following trails 4, 3, 2, and 1.

这题各种单源最短路都能过

#include<cstdio>
#include<cstring>
int n,m,x,y,z,cnt,t,w=1;
int head[10001];
struct edge{
int to,next,v;
}e[50001];
int q[50001];
bool mark[50001];
int dist[50001];
inline void ins(int u,int v,int w)
{
e[++cnt].v=w;
e[cnt].to=v;
e[cnt].next=head[u];
head[u]=cnt;
}
void insert(int u,int v,int w)
{
ins(u,v,w);
ins(v,u,w);
}
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline void spfa()
{
q[1]=1;
mark[1]=1;
dist[1]=0;
while (t<w)
{
int now=q[++t];
for (int i=head[now];i;i=e[i].next)
if (dist[now]+e[i].v<dist[e[i].to])
{
dist[e[i].to]=dist[now]+e[i].v;
if (!mark[e[i].to])q[++w]=e[i].to;
}
mark[now]=0;
}
}
int main()
{
m=read();n=read();
memset(dist,127/3,sizeof(dist));
for(int i=1;i<=m;i++)
{
x=read();y=read();z=read();
insert(x,y,z);
}
spfa();
printf("%d",dist[n]);
}

bzoj1752 [Usaco2005 qua]Til the Cows Come Home的更多相关文章

  1. BZOJ1754: [Usaco2005 qua]Bull Math

    1754: [Usaco2005 qua]Bull Math Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 374  Solved: 227[Submit ...

  2. 1753: [Usaco2005 qua]Who's in the Middle

    1753: [Usaco2005 qua]Who's in the Middle Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 290  Solved:  ...

  3. 1754: [Usaco2005 qua]Bull Math

    1754: [Usaco2005 qua]Bull Math Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 398  Solved: 242[Submit ...

  4. POJ 2387 Til the Cows Come Home(最短路 Dijkstra/spfa)

    传送门 Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 46727   Acce ...

  5. Til the Cows Come Home(最短路)

    Til the Cows Come Home Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I ...

  6. POJ2387 Til the Cows Come Home(SPFA + dijkstra + BallemFord 模板)

    Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37662   Accepted ...

  7. POJ 2387 Til the Cows Come Home

    题目链接:http://poj.org/problem?id=2387 Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K ...

  8. 怒学三算法 POJ 2387 Til the Cows Come Home (Bellman_Ford || Dijkstra || SPFA)

    Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 33015   Accepted ...

  9. POJ 2387 Til the Cows Come Home (最短路 dijkstra)

    Til the Cows Come Home 题目链接: http://acm.hust.edu.cn/vjudge/contest/66569#problem/A Description Bessi ...

随机推荐

  1. wordpress提速插件

    auto-remove-googles-url插件,替换前后台国外字体!访问速度有较大提高!可百度搜索auto-remove-googles-url下载,如在wp后台进行插件安装即可

  2. Struct2(五)处理表单

    简介: 1.表单的提交 表单和对应的Java模型的类 在此次的例子中,我们将会模仿一个用户提交表单的动作,具体提交表单做什么,不关心,我们需要知道 first last Name,Email addr ...

  3. python倒计时

    #coding=utf-8 #!/usr/bin/env python import datetime,time i=3 while i==3: spring=datetime.datetime(20 ...

  4. Java中普通代码块,构造代码块,静态代码块的代码演示样例及区分

    //运行顺序:(优先级从高到低.)静态代码块>mian方法>构造代码块>构造方法. 当中静态代码块仅仅运行一次.构造代码块在每次创建对象是都会运行. 1 普通代码块 <span ...

  5. 升级IOS9,提示滑动升级,卡在password锁屏界面,无反应了

    注:升级之前一定要把锁屏password取消掉 若遇上述问题.可通过进入DFU 模式解决 进入DFU具体步骤.(进入成功后,屏幕为全黑) 注:在进入DFU操作时务必与电脑连接好数据线. 1.按住pow ...

  6. JMeter脚本参数化和断言设置( CSV Data Set Config )

    用Badboy录制了Jmeter的脚本,用Jmeter打开后形成了原始的脚本.但是在实际应用中,为了增强脚本的多样性,就要使脚本参数化.这里我以登录为例,参数化用户账号与用户密码.  图1 :原始脚本 ...

  7. Sftp和ftp 差别、工作原理等(汇总ing)

    Sftp和ftp over ssh2的差别 近期使用SecureFx,涉及了两个不同的安全文件传输协议: -sftp -ftp over SSH2 这两种协议是不同的.sftp是ssh内含的协议,仅仅 ...

  8. OC教程10-NSNumber具体

    NSNumber简单介绍 NSNumber是数字的对象形式,由于在OC的数组和字典中仅仅同意存放对象,所以我们有时候须要转化 我们普通的类型是   123 那么 NSNumber类型的是  @123, ...

  9. C#基础:C#4.0权威指南 杂笔一

    1.c#中数组初始化的几种不同用法     int[] name = new int[NUM];       int[] name = {1, 2, 3, 4, 5, 6};       int[] ...

  10. 查看linux版本和内核信息

    一.查看Linux内核版本命令(两种方法): 1.cat /proc/version [root@localhost ~]# cat /proc/versionLinux version 2.6.32 ...