集训队专题(3)1009 Choose the best route
2016-02-05 01:27
309 查看
Choose the best route
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11546 Accepted Submission(s): 3751
Problem Description
One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s home so that she can take.
You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.
Input
There are several test cases.
Each case begins with three integers n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number of bus stations in this city and m stands for the number of directed ways between bus stations .(Maybe there are several ways between two bus stations .) s stands
for the bus station that near Kiki’s friend’s home.
Then follow m lines ,each line contains three integers p , q , t (0<t<=1000). means from station p to station q there is a way and it will costs t minutes .
Then a line with an integer w(0<w<n), means the number of stations Kiki can take at the beginning. Then follows w integers stands for these stations.
Output
The output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.
Sample Input
5 8 5 1 2 2 1 5 3 1 3 4 2 4 7 2 5 6 2 3 5 3 5 1 4 5 1 2 2 3 4 3 4 1 2 3 1 3 4 2 3 2 1 1
Sample Output
1 -1
Author
dandelion
Source
2009浙江大学计算机研考复试(机试部分)——全真模拟
题意:给出n,m,s,代表有n个车站,m条路线,s为终点然后m行代表路线和花费,注意是单向边。最后给出k个数代表能选择的出发点。
思路:这道题小编觉得其实和一个人的旅行的题目大同小异,将0看成出发点,将n个能选择的点看做事0的相邻并且花费为0的点,那么就是求0到S的最小花费了。但是……最大的不同是,此题的数据较大,我们不能用Floyd算法(小编TT),还是用dijkstra算法吧。
#include <stdio.h> #include <string.h> #include <algorithm> #define INF 0xfffffff using namespace std; int n,m,s; int map[1005][1005]; int vis[1005],cast[1005]; void Dijkstra() { int i,j,minn,pos; memset(vis,0,sizeof(vis)); vis[0] = 1; for(i = 0; i<=n; i++) cast[i] = map[0][i]; for(i = 0; i<=n; i++) { minn = INF; for(j = 0; j<=n; j++) { if(cast[j]<minn && !vis[j]) { pos = j; minn = cast[j]; } } vis[pos] = 1; for(j = 0; j<=n; j++) { if(cast[pos]+map[pos][j]<cast[j] && !vis[j]) cast[j] = cast[pos]+map[pos][j]; } } } int main() { int i,j; int x,y,t; while(~scanf("%d%d%d",&n,&m,&s)) { for(i = 0; i<=n; i++) for(j = 0; j<=n; j++) map[i][j] = INF; while(m--) { scanf("%d%d%d",&x,&y,&t); if(t<map[x][y]) map[x][y] = t; } scanf("%d",&m); while(m--) { scanf("%d",&x); map[0][x] = 0; } Dijkstra(); if(cast[s] == INF) printf("-1\n"); else printf("%d\n",cast[s]); } return 0; }
相关文章推荐
- iOS客户端的APNS服务简介与实现
- 集训队专题(3)1008 最短路
- Quartz 定时器任务调度
- 集训队专题(3)1007 HDU Today
- 【BZOJ1082】【codevs2456】栅栏,让人抓狂的优化剪枝
- algorithm@ lower_bound implementation(Binary Search)
- 集训队专题(3)1006 一个人的旅行
- Linux下安装配置Nexus
- [工作中的设计模式]备忘录模式memento
- 集训队专题(3)1005 畅通工程续
- ruby on rails爬坑(二):Active Record数据库查询
- 【HDOJ】3553 Just a String
- 集训队专题(3)1004 六度分离
- hdu 1829 种类并查集
- Mosquitto安装、测试总结Fedora14
- Android基础之Fragment
- PAT 1003 Emergency
- 使用Xcode7报App Transport Security has blocked a cleartext HTTP错误
- 分解数字
- React 从0开始 消息传递