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.
題意:給出n,m,s,代表有n個車站,m條路線,s爲終點,然後m行代表路線和花費,單向邊,最後給出k個數代表能選擇的出發點
思路:Dijkstra算法;
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int inf = 100000000;
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;
}