hdu 2680：Choose the best route（Dijkstra , SPFA）

http://acm.hdu.edu.cn/showproblem.php?pid=2680

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

题意分析：

有n个车站，可以选取w个中任意一个作为起点，车站之间有单向通行时间，求到达终点的最短时间。

解题思路：

因为有多个起点，把终点当作起点，起点当终点， 反向建边，求出最短路后比较所有终点的值。

注意有重复边。

Dijkstra算法：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 1020
using namespace std;
int n, m, s, w, inf=0x3f3f3f3f;
int a[N], e[N][N], dis[N];
bool book[N];
void init()
{
	int i, j, u, v, c;
	for(i=1; i<=n; i++)
		for(j=1; j<=n; j++)
			if(i==j)
				e[i][j]=0;
			else e[i][j]=inf;
	for(i=1; i<=m; i++){
		scanf("%d%d%d", &u, &v, &c);
		if(e[v][u]>c)
			e[v][u]=c;
	}
	scanf("%d", &w);
	for(i=1; i<=w; i++)
		scanf("%d", &a[i]);		
}
int Dijkstra()
{
	int mini, u;
	memset(book, false, sizeof(book));
	book[s]=true;
	for(int i=1; i<=n; i++)
		dis[i]=e[s][i];
	for(int i=1; i<n; i++)
	{
		mini=inf;
		for(int j=1; j<=n; j++){
			if(book[j]==false && dis[j]<=mini){
				mini=dis[j];
				u=j;
			}	
		}	
		book[u]=true;
		for (int v=1; v<=n; v++)
			if (book[v]==false && dis[v]>dis[u]+e[u][v])
				dis[v]=dis[u]+e[u][v]; 
	} 
	int ans=inf;
	for(int i=1; i<=w; i++)
		ans=min(ans, dis[a[i]]);
	if(ans==inf)
		return -1;
	return ans;
}
int main()
{
	while(scanf("%d%d%d", &n, &m, &s)!=EOF)
	{
		init();
		printf("%d\n", Dijkstra());
	}
	return 0;
} 

SPFA：

#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <algorithm>
#define N 1020
using namespace std;
int n, m, s, w, inf=0x3f3f3f;
int a[N], dis[N];
bool book[N];
struct edge{
	int v;
	int w;
	edge(int v, int w) : v(v), w(w) {}
};
vector <edge>e[N];
void init()
{
	int i, j, u, v, c;
	for(i=1; i<=n; i++)
		e[i].clear();
	for (i=1; i<=n; i++)
		dis[i]=inf;
	dis[s]=0; 
	for (i=1; i<=m; i++){
		scanf("%d%d%d", &u, &v, &c);
		e[v].push_back(edge(u, c));
	}
	scanf("%d", &w);
	for (i=1; i<=w; i++)
		scanf("%d", &a[i]);		
}
int SPFA()
{
	int mini, u;
	queue<int>q;
	memset(book, false, sizeof(book));
	q.push(s);
	book[s]=true;
	while(!q.empty())
	{
		u=q.front();
		q.pop();
		book[u]=false;
		for(int i=0; i<e[u].size(); i++){
			if(dis[e[u][i].v]>dis[u]+e[u][i].w)
			{
				dis[e[u][i].v]=dis[u]+e[u][i].w;
				if(book[e[u][i].v] == false)
				{
					q.push(e[u][i].v);
					book[e[u][i].v]=true;
				} 
			}
		}
	}
	int ans=inf;
	for(int i=1; i<=w; i++)
		ans=min(ans, dis[a[i]]);
	if(ans==inf)
		return -1;
	return ans;
}
int main()
{
	while(scanf("%d%d%d", &n, &m, &s)!=EOF)
	{
		init();
		printf("%d\n", SPFA());
	}
	return 0;
}