Choose the best route
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12278 Accepted Submission(s):
3986
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.
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
Recommend
很简单的迪杰斯特拉，和以前做的一道题是一样的，因为有多个起点，所以将0作为起点，到所有原本起点的距离为0即可。
题意：给出n个站，m条路（有向）和终点站。输入m条点到点的距离，再给出q，表示可作为起点的站编号，给出这q个编号，求最短路。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define inf 0x3f3f3f3f 5 #define M 1005 6 using namespace std; 7 int map[M][M],vis[M],dis[M]; 8 int main() 9 { 10 int i,j,n,m,s; 11 while(~scanf("%d%d%d",&n,&m,&s)) 12 { 13 memset(vis,0,sizeof(vis)); 14 memset(dis,0,sizeof(dis)); 15 for(i=0; i<=n; i++) 16 for(j=0; j<=n; j++) 17 { 18 if(i==j) map[i][j]=0; 19 else map[i][j]=inf; 20 } 21 int a,b,c; 22 for(i=0; i<m; i++) 23 { 24 scanf("%d%d%d",&a,&b,&c); 25 if(map[a][b]>c) 26 map[a][b]=c; 27 } 28 int w,q; 29 scanf("%d",&w); 30 for(i=0; i<w; i++) 31 { 32 scanf("%d",&q); 33 map[0][q]=0; //将0作为起点！ 34 } 35 for(i=0; i<=n; i++) 36 dis[i]=map[0][i]; 37 vis[0]=1; 38 int min,t; 39 for(i=1; i<=n; i++) 40 { 41 min=inf; 42 for(j=1; j<=n; j++) 43 if(!vis[j]&&min>dis[j]) 44 { 45 min=dis[j]; 46 t=j; 47 } 48 vis[t]=1; 49 for(j=1; j<=n; j++) 50 if(!vis[j]&&map[t][j]<inf) 51 if(dis[j]>dis[t]+map[t][j]) 52 dis[j]=dis[t]+map[t][j]; 53 } 54 if(dis[s]<inf) 55 printf("%d ",dis[s]); 56 else 57 printf("-1 "); 58 } 59 return 0; 60 }