HDU 2680(SPFA + 反向)

Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17549    Accepted Submission(s): 5684

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

 

Author

dandelion

题目分析：    题目大意很简单，就是一些路径，给你一些起点，一个终点，每个路径有不同的权重，然后让你在这些起点中挑出一个起点，使得从这个起点到终点的距离最短。如果每一个起点都进行一遍spfa的话，会造成超时，所以我们把终点当做起点，把起点当做终点，这样一来，一遍SPFA就可以搞定了。但是要注意，存储路径的时候，要反着存，因为这个图已经是反的了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int inf = 99999999;
const int maxn = 1005;
int map[maxn][maxn];
long long dis[maxn];
int vis[maxn],q[maxn];

void init()
{
	memset(vis,0,sizeof(vis));
	memset(dis,0,sizeof(dis));
	memset(map,0,sizeof(map));
	memset(q,0,sizeof(q));
	
	for(int i=0; i<=maxn;i++)
		for(int j=0;j<=maxn;j++)
			if(i != j)
				map[i][j] = inf;
}
void spfa(int s,int n)
{
	for(int i = 1; i <= n; i++)
		dis[i] = inf;
	
	dis[s] = 0; vis[s] = 1;
	q[1] = s;
	int v,head = 0,tail = 1;
	
	while(head < tail)
	{
		head++;
		v = q[head];
		vis[v] = 0;
		for(int i = 1; i <= n; i++)
		{
			if(map[v][i] > 0 && dis[i] > dis[v]+map[v][i])
			{
				dis[i] = dis[v]+map[v][i];
				if(vis[i] == 0)
				{
					vis[i] = 1;
					tail++;
					q[tail] = i;
				}
			}
			
		}
	 } 
}
int main()
{
	int n,m,s;
	
	while((scanf("%d%d%d",&n,&m,&s))!=EOF)
	{
		init();
		while(m--)
		{
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			if(map[b][a] > c)
				map[b][a] = c;
		}
		int w;
		int mn = inf;
		int ss[maxn]={0};
		scanf("%d",&w);
		for(int i = 1; i <= w; i++)
			scanf("%d",&ss[i]);
			
		spfa(s,n);
		
		for(int i = 1; i <= w; i++)
			if(mn > dis[ss[i]])
				mn = dis[ss[i]];

		if(mn == inf)	cout<<"-1"<<endl;
		else cout<<mn<<endl;
	}
	return 0;
 }