博客
关于我
POJ 2387 Til the Cows Come Home(Dijkstra优先队列)
阅读量:331 次
发布时间:2019-03-04

本文共 2391 字,大约阅读时间需要 7 分钟。

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
Hint
INPUT DETAILS:
There are five landmarks.
OUTPUT DETAILS:
Bessie can get home by following trails 4, 3, 2, and 1.

优先队列定义q就直接一个结构体就可以了,不要在写vector还是greater或者less了,写上去还麻烦,真的是伤心了。

#include 
#include
#include
#include
using namespace std;const int inf = 0x3f3f3f3f;struct node{ int d; int pos; };priority_queue
q;bool operator < (node a, node b){ return a.d > b.d;//权值小的边优先}int n , m;int e[2005][2005];int dis[2005];void dijkstra(int start){ int book[2005]; memset(book, 0, sizeof(book)); for (int i=1; i<=n; i++) if (i==start) dis[i]=0; else dis[i]=inf; node t; t.d=dis[start]; t.pos=start; q.push(t); while(!q.empty()){ t = q.top(); q.pop(); int x = t.pos; if(book[x]==1) continue; book[x] = 1; for(int j=1; j<=n; j++){ if(book[j]==0 && dis[j]>dis[x]+e[x][j]){ dis[j]=dis[x]+e[x][j]; node tt; tt.d=dis[j]; tt.pos=j; q.push(tt); } } } cout<
<
w){ e[u][v]=w; e[v][u]=w; } } dijkstra(1); } return 0;}

转载地址:http://lpnh.baihongyu.com/

你可能感兴趣的文章
mysql大批量删除(修改)The total number of locks exceeds the lock table size 错误的解决办法
查看>>
mysql如何做到存在就更新不存就插入_MySQL 索引及优化实战(二)
查看>>
mysql如何删除数据表,被关联的数据表如何删除呢
查看>>
MySQL如何实现ACID ?
查看>>
mysql如何记录数据库响应时间
查看>>
MySQL子查询
查看>>
Mysql字段、索引操作
查看>>
mysql字段的细节(查询自定义的字段[意义-行列转置];UNION ALL;case-when)
查看>>
mysql字段类型不一致导致的索引失效
查看>>
mysql字段类型介绍
查看>>
mysql字段解析逗号分割_MySQL逗号分割字段的行列转换技巧
查看>>
MySQL字符集与排序规则
查看>>
MySQL字符集乱码
查看>>
mysql字符集设置
查看>>
mysql存储IP地址的数据类型
查看>>
mysql存储中文 但是读取乱码_mysql存储中文乱码
查看>>
MySQL存储引擎
查看>>
MySQL存储引擎
查看>>
MySQL存储引擎--MYSIAM和INNODB引擎区别
查看>>
Mysql存储引擎(1):存储引擎体系结构和介绍
查看>>