HDOJ 2680 Choose the best route(最短路—反向图，dijkstra算法)

Choose the best route

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 10600 Accepted Submission(s): 3414



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

题意：我觉得大家都看得懂，因为我看懂了。

解题思路：终点站只有一个，起点却又多种选择。所以可以建立反向图，将终点站当做图的起点，起点站设为图的终点。
另外注意题中给出的路径都是单向的。

dijkstra算法，代码如下：

<span style="font-size:12px;">#include<cstdio>
#include<cstring>
#define INF 0x3f3f3f
#define max 1010
int n,map[max][max];
int dis[max];

void dijkstra(int start)
{
	int i,j,min,next=start;
	int visit[max];
	memset(visit,0,sizeof(visit));
	for(i=1;i<=n;++i)
	   dis[i]=map[start][i];
	visit[start]=1;
	for(i=1;i<=n;++i)
	{
		min=INF;
		for(j=1;j<=n;++j)
		{
			if(!visit[j]&&dis[j]<min)
			{
				min=dis[j];
				next=j;
			}
		}
		visit[next]=j;
		for(j=1;j<=n;++j)
		{
			if(!visit[j]&&dis[j]>dis[next]+map[next][j])
			   dis[j]=dis[next]+map[next][j];
		}
	}
} 

int main()
{
	int i,j,m,s,a,b,c,sta,w;
	while(scanf("%d%d%d",&n,&m,&s)!=EOF)
	{
		for(i=1;i<=n;++i)
		{
			for(j=1;j<=n;++j)
			{
				if(i==j)
				   map[i][j]=0;
				else
				   map[i][j]=INF;
			}
		}
		while(m--)
		{
			scanf("%d%d%d",&a,&b,&c);//反向图，即从b到a。 
			if(map[b][a]>c)
			   map[b][a]=c;//hdu的路都是单行道，有木有很无语 
		}
		dijkstra(s);
		scanf("%d",&w);
		int ans=INF;
		while(w--)
		{
			scanf("%d",&sta);
			if(dis[sta]<ans)
			   ans=dis[sta];
		}
		if(ans==INF)
		   printf("-1\n");
		else
		   printf("%d\n",ans);
	}
	return 0;
}</span>