hdoj 2680 最短路(dijkstra)
2016-02-02 01:05
309 查看
Choose the best route
[align=left]Problem Description[/align]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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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”.
[align=left]Sample Input[/align]
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
[align=left]Sample Output[/align]
1 -1
[align=left]Author[/align]
dandelion
[align=left]Source[/align]
2009浙江大学计算机研考复试(机试部分)——全真模拟
找了2小时的错,最后我发现
路是单向的!!!
牢记教训!!!
起点不确定,终点确定,所以我们可以把终点当成起点,反着进行寻路过程,反正一样的- -,这样只要进行一次dijkstra就行
#include<cstdio> #include<cstring> using namespace std; #define INF 0x3f3f3f3f const int MAX=1010; int dist[MAX],n,map[MAX][MAX],vis[MAX]; void Dijkstra(int s) { int i,j,min,pos; memset(vis,0,sizeof(vis)); for(i=1;i<=n;i++) dist[i]=map[s][i]; dist[s]=0; vis[s]=1; for(i=1;i<=n;i++) { min=INF; for(j=1;j<=n;j++) { if(!vis[j]&&dist[j]<min) min=dist[pos=j]; } if(min==INF) break; vis[pos]=1; for(j=1;j<=n;j++) { if(!vis[j]&&dist[pos]+map[pos][j]<dist[j]) dist[j]=dist[pos]+map[pos][j]; } } } int main() { int i,e,a,b,c,x,y,m; while(scanf("%d%d%d",&n,&m,&e)!=EOF) { memset(map,INF,sizeof(map)); for(i=1;i<=m;i++) { scanf("%d%d%d",&a,&b,&c); if(c<map[b][a]) map[b][a]=c; } Dijkstra(e); scanf("%d",&x); int ans=INF; while(x--) { scanf("%d",&y); if(dist[y]<ans) ans=dist[y]; } if(ans!=INF) printf("%d\n",ans); else printf("-1\n"); } return 0; }
相关文章推荐
- mybatis教程
- HUST 1017 Exact cover (精确覆盖|Dancing Links模板题)
- Remove Duplicate Letters
- 心态不好,你注定是个弱者!
- iOS知识小集 第一期
- 网页鼠标滚轮事件(滚滚屏)的获取及理解
- 【程序员在法国】两记棒喝,惊醒梦中人
- 安装kali linux后遇到的问题解决方案
- 解决thinkpad维持“正在关机”及“启动PWMTR64V.DLL时出现问题”
- C#流使用总结
- 【程序员在法国】两记棒喝,惊醒梦中人
- 树莓派:VNC远程登录Raspbian图形界面(tightvncserver)
- 开发明细原创、问题和hibernate3与4的区别转载
- 多线程中的join()方法
- HDD-FAT32 ZIP-FAT32
- redis原理
- 【jQuery基础学习】06 jQuery表单验证插件-Validation
- ubuntu中vmware出现:Network configuration is missing. Ensure that /etc/vmware/networking exists.的处理方式
- JAVA_安装JDK和Eclipse
- 分布式缓存系统 Memcached 整体架构