Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18567    Accepted Submission(s): 6011

 

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

 

 

Source

2009浙江大学计算机研考复试（机试部分）——全真模拟

 

 

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      n个车站，m条途径，多个起点和一个终点，没什么可以说的，就是要注意设置一个超级源点0，在最后把他跟所有的起点连一个长度为0的边就行；

      不过要注意的是。。。。这道题中边是个单向边。。。；

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
struct ***
{
	int to, len, ne;
}ed[50005];
int head[1005];
int d[1005];
bool vis[1005];
int cnt = 0, n, m, st, en;
void init()
{
	cnt = 0;
	for (int s = 0; s <= n; s++)
	{
		head[s] = -1;
		vis[s] = 0;
		d[s] = 0x3f3f3f3f;
	}
}
void add(int from, int to, int len)
{
	ed[cnt].to = to;
	ed[cnt].ne = head[from];
	ed[cnt].len = len;
	head[from] = cnt++;
}
void spfa()
{
	queue<int>q;
	q.push(0);
	vis[0] = 1;
	d[0] = 0;
	while (!q.empty())
	{
		int t = q.front();
		q.pop();
		vis[t] = 0;
		for (int s = head[t]; ~s; s = ed[s].ne)
		{
		//	cout << s << endl;
			if (d[ed[s].to] > d[t] + ed[s].len)
			{
				d[ed[s].to] = d[t] + ed[s].len;
				if (!vis[ed[s].to])
				{
					q.push(ed[s].to);
					vis[ed[s].to] = 1;
				}
			}
		}
	}
}
int main()
{
	while (~scanf("%d%d%d", &n, &m, &en))
	{
		init();
		while (m--)
		{
			int a, b, c;
			scanf("%d%d%d", &a, &b, &c);
			add(a, b, c);
		}
		int sum;
		cin >> sum;
		while (sum--)
		{
			scanf("%d", &st);
			add(0, st, 0);
		}
		spfa();
		if (d[en] == 0x3f3f3f3f)
		{
			printf("-1\n");
		}
		else
		{
			printf("%d\n", d[en]);
		}
	}
	return 0;
}