POJ2387 Til the Cows Come Home(SPFA + dijkstra + BallemFord 模板)
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 37662 | Accepted: 12836 |
Description
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
* 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
Sample Input
5 5 1 2 20
3 4 20
4 5 20
2 3 30 1 5 100
Sample Output
90
Hint
There are five landmarks.
OUTPUT DETAILS:
Bessie can get home by following trails 4, 3, 2, and 1.
SPFA:
#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
#include <cstring>
#include <algorithm> using namespace std;
const int INF = ;
const int MAX = + ;
int t,n;
struct point
{
int e,w;
};
vector<point> g[MAX];
int dist[MAX];
void spfa(int v)
{
for(int i = ; i <= n; i++)
{
dist[i] = INF;
}
dist[v] = ;
queue<int> que;
que.push(v);
while(que.size())
{
int x = que.front();
que.pop();
int len = g[x].size();
for(int i = ; i < len; i++)
{
int y = g[x][i].e;
if(dist[y] > dist[x] + g[x][i].w)
{
dist[y] = dist[x] + g[x][i].w;
que.push(y);
}
}
}
}
int main()
{
while(scanf("%d%d", &t, &n) != EOF)
{
for(int i = ; i < MAX; i++)
g[i].clear(); while(t--)
{
int s,e,w;
point temp;
scanf("%d%d%d", &s,&e,&w);
temp.w = w;
temp.e = e;
g[s].push_back(temp);
temp.e = s;
g[e].push_back(temp);
} spfa(n);
printf("%d\n",dist[]);
} return ;
}
SPFA
Dijkstra
注意重边问题
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <string.h>
using namespace std;
const int INF = ;
const int MAX = + ;
int g[MAX][MAX],dist[MAX],vis[MAX];
int t,n;
void Dijkstra()
{
for(int i = ; i <= n; i++)
dist[i] = INF;
memset(vis,,sizeof(vis));
dist[n] = ;
vis[n] = ;
int pos = n;
for(int i = ; i < n; i++)
{
int minn = INF;
for(int j = ; j <= n; j++)
{
if(vis[j] == && dist[j] < minn)
{
minn = dist[j];
pos = j;
}
}
vis[pos] = ;
for(int j = ; j <= n; j ++)
{
if(vis[j] == && dist[j] > dist[pos] + g[pos][j])
dist[j] = dist[pos] + g[pos][j];
}
}
}
int main()
{
while(scanf("%d%d",&t,&n) != EOF)
{
int s,e,w;
for(int i = ; i <= n; i++)
{
for(int j = ; j <= n; j++)
{
g[i][j] = INF;
}
}
for(int i = ; i < t; i++)
{
scanf("%d%d%d",&s,&e,&w);
if(g[s][e] > w)
g[s][e] = g[e][s] = w;
}
Dijkstra();
printf("%d\n",dist[]);
} }
Ballem_ford
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
const int INF = ;
const int MAX = + ;
int n,t;
struct point
{
int s,t,w;
};
vector<point> g;
int dist[MAX];
void Ballem_ford(int v)
{
for(int i = ; i <= n; i++)
dist[i] = INF;
dist[v] = ;
for(int j = ; j < n; j++)
{
int len = g.size();
int flag = ;
for(int i = ; i < len; i++)
{
int s = g[i].s;
int t = g[i].t;
int w = g[i].w;
if(dist[t] > dist[s] + w)
{
dist[t] = dist[s] + w;
flag = ;
}
}
if(flag == ) //加个flag 优化一下
break;
}
}
int main()
{
while(scanf("%d%d", &t, &n) != EOF)
{
g.clear();
int s,e,w;
point temp;
for(int i = ; i < t; i++)
{
scanf("%d%d%d", &s,&e,&w);
temp.w = w;
temp.t = e;
temp.s = s;
g.push_back(temp);
temp.t = s;
temp.s = e;
g.push_back(temp);
}
Ballem_ford(n);
printf("%d\n",dist[]);
}
}
POJ2387 Til the Cows Come Home(SPFA + dijkstra + BallemFord 模板)的更多相关文章
- poj2387 Til the Cows Come Home 最短路径dijkstra算法
Description Bessie is out in the field and wants to get back to the barn to get as much sleep as pos ...
- poj2387 Til the Cows Come Home(Dijkstra)
题目链接 http://poj.org/problem?id=2387 题意 有n个路标,编号1~n,输入路标编号及路标之间相隔的距离,求从路标n到路标1的最短路径(由于是无向图,所以也就是求从路标1 ...
- POJ2387 Til the Cows Come Home 【Dijkstra】
题目链接:http://poj.org/problem?id=2387 题目大意; 题意:给出两个整数T,N,然后输入一些点直接的距离,求N和1之间的最短距离.. 思路:dijkstra求单源最短路, ...
- POJ2387 Til the Cows Come Home (最短路 dijkstra)
AC代码 POJ2387 Til the Cows Come Home Bessie is out in the field and wants to get back to the barn to ...
- POJ.2387 Til the Cows Come Home (SPFA)
POJ.2387 Til the Cows Come Home (SPFA) 题意分析 首先给出T和N,T代表边的数量,N代表图中点的数量 图中边是双向边,并不清楚是否有重边,我按有重边写的. 直接跑 ...
- 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 ...
- (Dijkstra) POJ2387 Til the Cows Come Home
Til the Cows Come Home Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 81024 Accepted ...
- POj2387——Til the Cows Come Home——————【最短路】
A - Til the Cows Come Home Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & ...
- POJ-2387.Til the Cows Come Home.(五种方法:Dijkstra + Dijkstra堆优化 + Bellman-Ford + SPFA + Floyd-Warshall)
昨天刚学习完最短路的算法,今天开始练题发现我是真的菜呀,居然能忘记邻接表是怎么写的,真的是菜的真实...... 为了弥补自己的菜,我决定这道题我就要用五种办法写出,并在Dijkstra算法堆优化中另外 ...
随机推荐
- .Net core环境准备
.Net core 出来有段日子了,在跨平台上迈出了坚实的一步,尽管如此身边还是有很多人都转向了Java阵营.抛开语言之争,在.net平台上工作多年,还是有必要了解下新推出的技术,没准有朝一日就用上了 ...
- f2fs解析(四)f2fs的extent特性
extent的意思是“程度”,但是我还是搞不清楚要如何把“程度”和我理解的extent联系到一起. 文件的偏移和page-cache的映射关系体现在address space 中的一颗基数树上:当基数 ...
- 第三章 Models模块属性详解
摘自:http://www.cnblogs.com/xdotnet/archive/2012/03/07/aspnet_mvc40_validate.html 了解了这些就可以对MVC进一步认识,相信 ...
- Linux 信号概念
程序在执行的时候,几乎任何时刻都会反生事件. 信号通常用来向一个进程通知事件. 信号是不可提前预知的,所以信号是异步的. 信号随时都可能发生,接收信号的进程也可以没有控制权. 每个信号名都以SIG开头 ...
- CentOS7 SSH相关
1.查看SSH是否已经安装,命令: rpm -qa |grep ssh 如果列出了openssh-x.x开头的软件名,代表已经安装 如果没有安装,使用命令安装: yum install ssh 2.如 ...
- JS 之性能优化(1)
了解JS性能优化是学习前端必备的一项技能.下面就简单的列出几点: 1.注意作用域,避免全局查找. 访问全局变量比访问局部变量慢,是因为需要遍历作用域链,查找作用域链需要额外的时间.所以在一个函数中,将 ...
- sql server 2008 登录 4064 错误解决办法
出现这个错误是账户无法打开默认数据库导致的 修改一下该账户的默认打开数据库即可. 如果很不幸,你的sa帐号和windows身份验证默认都打开同一个数据库,那么无论换哪种方式登录都是一样没用的 如果你有 ...
- php 利用socket发送GET,POST请求
作为php程序员一定会接触http协议,也只有深入了解http协议,编程水平才会更进一步.最近我一直在学习php的关于http的编程,许多东西恍然大悟,受益匪浅.希望分享给大家.本文需要有一定http ...
- #Linux学习笔记# Linux文件的所有者、群组和其他人
1.关于所有者.群组和其他人 在Linux系统中,每个文件都具有User.Group和Others三种身份的权限配置.那这三种身份分别表示什么意思呢?配置这三种身份的权限有啥意义呢? (1)文件所有者 ...
- Android新组件RecyclerView介绍,其效率更好
今天我们首先来说为什么要介绍这个新组件RecyclerView,因为前几天我发布了一个常用面试题ListView的复用及如何优化的文章,介绍给一些开发者,但是我看到有关的反馈说:现在都不再用listv ...