Choose the best route--hdu2680
Choose the best route
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10482 Accepted Submission(s): 3373
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.
很明显的要用dijkstra,但是估计大家用一般的方法建图的时候都会超时吧,因为起点不止一个,起点多就要多次调用函数,因此超时!
倒不如反向建图,变成一个起点,多个终点!
另外注意:公车车是单向的!
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAX 0x3f3f3f3f
using namespace std;
int map[][],d[],n,m,s,t;
void dijkstra(int x)
{
int i,j,min,mark,used[];
for(i=;i<=n;i++)
{
used[i]=;
d[i]=map[x][i];
}
d[x]=;
used[x]=;
for(i=;i<=n;i++)
{
min=MAX;
mark=-;
for(j=;j<=n;j++)
{
if(!used[j]&&d[j]<min)
{
min=d[j];
mark=j;
}
}
if(mark==-)
break;
used[mark]=;
for(j=;j<=n;j++)
{
if(!used[j]&&d[j]>d[mark]+map[mark][j])
d[j]=d[mark]+map[mark][j];
}
} }
int main()
{
int a,b,c,i,j;
while(scanf("%d%d%d",&n,&m,&s)!=EOF)
{
memset(map,MAX,sizeof(map));
for(i=;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
if(map[b][a]>c)
map[b][a]=c;
}
dijkstra(s);//调用一次
int start;
int mi=MAX;
scanf("%d",&t);
for(i=;i<t;i++)//找出最小的花费
{
scanf("%d",&start);
mi=mi<d[start]?mi:d[start];
}
if(mi==MAX)
printf("-1\n");
else
printf("%d\n",mi);
}
return ;
}
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