POJ 2387 Til the Cows Come Home Dijkstra求最短路径
Til the Cows Come Home
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
2 3 30
3 4 20
4 5 20
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.
#include<stdio.h>
#include<string.h>
#include<limits.h> int a[][];
int dis[],b[];
int n; int min(int x,int y)
{
return x<y?x:y;
} void dij(int k)
{
int i,j,mind,minj;
memset(b,,sizeof(b));
for(i=;i<=n;i++){
dis[i]=INT_MAX;
}
dis[k]=;
for(i=;i<n;i++){
mind=INT_MAX;
for(j=;j<=n;j++){
if(!b[j]&&dis[j]<mind){
mind=dis[j];
minj=j;
}
}
b[minj]=;
for(j=;j<=n;j++){
if(!b[j]&&a[minj][j]>=){
dis[j]=min(dis[j],dis[minj]+a[minj][j]);
}
}
}
} int main()
{
int t,x,y,z;
memset(a,-,sizeof(a));
scanf("%d%d",&t,&n);
while(t--){
scanf("%d%d%d",&x,&y,&z);
if(a[x][y]!=-){
if(z<a[x][y]){
a[x][y]=z;
a[y][x]=z;
}
}
else{
a[x][y]=z;
a[y][x]=z;
}
}
dij();
printf("%d\n",dis[n]);
return ;
}
优化Dij:
#include<stdio.h>
#include<string.h>
#include<limits.h>
#include<vector>
#include<queue>
using namespace std; struct Node{
int v,w;
friend bool operator<(Node a,Node b)
{
return a.w>b.w;
}
}node; vector<Node> a[];
int dis[];
int n; void dij(int k)
{
int v1,v2,i;
priority_queue<Node> q;
dis[k]=;
node.w=;
node.v=k;
q.push(node);
while(q.size()){
v1=q.top().v;
q.pop();
for(i=;i<a[v1].size();i++){
v2=a[v1][i].v;
if(dis[v2]>dis[v1]+a[v1][i].w){
dis[v2]=dis[v1]+a[v1][i].w;
node.w=dis[v2];
node.v=v2;
q.push(node);
}
}
}
} int main()
{
int t,x,y,z,i;
scanf("%d%d",&t,&n);
for(i=;i<=n;i++){
a[i].clear();
dis[i]=INT_MAX;
}
while(t--){
scanf("%d%d%d",&x,&y,&z);
node.w=z;
node.v=y;
a[x].push_back(node);
node.v=x;
a[y].push_back(node);
}
dij();
printf("%d\n",dis[n]);
return ;
}
POJ 2387 Til the Cows Come Home Dijkstra求最短路径的更多相关文章
- Poj 2387 Til the Cows Come Home(Dijkstra 最短路径)
题目:从节点N到节点1的求最短路径. 分析:这道题陷阱比较多,首先是输入的数据,第一个是表示路径条数,第二个是表示节点数量,在 这里WA了四次.再有就是多重边,要取最小值.最后就是路径的长度的最大值不 ...
- poj 2387 Til the Cows Come Home(dijkstra算法)
题目链接:http://poj.org/problem?id=2387 题目大意:起点一定是1,终点给出,然后求出1到所给点的最短路径. 注意的是先输入边,在输入的顶点数,不要弄反哦~~~ #incl ...
- POJ 2387 Til the Cows Come Home (Dijkstra)
传送门:http://poj.org/problem?id=2387 题目大意: 给定无向图,要求输出从点n到点1的最短路径. 注意有重边,要取最小的. 水题..对于无向图,从1到n和n到1是一样的. ...
- POJ 2387 Til the Cows Come Home (图论,最短路径)
POJ 2387 Til the Cows Come Home (图论,最短路径) Description Bessie is out in the field and wants to get ba ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 九度OJ 1025:最大报销额 (01背包、DP)
时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:4352 解决:1055 题目描述: 现有一笔经费可以报销一定额度的发票.允许报销的发票类型包括买图书(A类).文具(B类).差旅(C ...
- 使用Fiddler测试HTTP接口
Fiddler下载地址:https://www.telerik.com/download/fiddler/fiddler4 在测试http接口前,为避免干扰,我们启用过滤器 然后运行过滤器设置 我们以 ...
- Django框架ORM常用参数汇总_模型层
primary_key 如果为True,那么这个字段就是模型的主键. 如果你没有指定任何一个字段的primary_key=True, Django就会自动添加一个IntegerField字段做为主键, ...
- PHP Framework
PHP Framework is built for PHP developers who need elegant toolkit to create full-featured web appli ...
- ScrollView当显示超出当前页面时自动移动到最底端【转】
本文转载自:http://gundumw100.iteye.com/blog/1162964 卷轴视图(ScrollView)是指当拥有很多内容,一屏显示不完时,需要通过滚动来显示视图.比如在做一个阅 ...
- 高通MSM8255 GPS 调试分析&&Android系统之Broadcom GPS 移植【转】
本文转载自:http://blog.csdn.net/gabbzang/article/details/12063031 http://blog.csdn.NET/dwyane_zhang/artic ...
- windows下的git使用指令
经常使用mac和linux 这次使用window开发了一些小项目 废话不多说: git init git add . git commit -m 'note' git remote add origi ...
- ubuntu14开发环境配置
1 配置JDK1.8 jdk工具从官网下载,我下载到了~/tool目录下,首先进入用户的bash配置目录,打开配置文件: cd ~ vi .bashrc 编辑.bashrc文件,在适当位置或者文件最后 ...
- HDU 4652 Dice:期望dp(成环)【错位相减】
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4652 题意: 给你一个有m个面的骰子. 两种询问: (1)"0 m n": “最后 ...
- CentOS 7编译安装Tengine+PHP+MariaDB全程笔记
安装环境:CentOS7 3.10.0-693.5.2.el7.x86_64 准备源码包: pcre-8.41.tar.gz openssl-1.0.1h.tar.gz zlib-1.2.11.tar ...