杭电2680-Choose the best route(最短路反向建图)

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本文介绍如何使用Dijkstra算法解决在给定城市公交路线和起始点的情况下，找到到达目标地点的最短路径的问题。通过实例分析，详细解释了算法的应用过程，包括初始化、松弛操作和更新最短路径的过程。特别强调了反向建图的技巧以适应题目条件，最终输出最少所需时间或不可达情况。

Choose the best route

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

Total Submission(s): 10891 Accepted Submission(s): 3529

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

Sample Output

1
-1

这个题用spfa没法判断重边，用floyd会超时，所以只能用dijkstra了，但是，记得要反向建图，因为只有一个终点，多个起点，所以反向建图只需要用一次dijlstra

#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int map[1010][1010],dis[1010],vis[1010];
int m,n,e,s[1010];
void dijkstra()
{
	int i,j,mi,mark;
	memset(vis,0,sizeof(vis));
	for(i=1;i<=m;i++)
		dis[i]=map[e][i];
	vis[e]=1;
	dis[e]=0;
	for(i=0;i<m;i++)
	{
		mi=INF,mark=-1;
		for(j=1;j<=m;j++)
		{
			if(mi>dis[j]&&!vis[j])
			{
				mi=dis[j];
				mark=j;
			}
		}
		if(mark==-1)
		break;
		vis[mark]=1;
		for(j=1;j<=m;j++)
		{
			if(!vis[j]&&dis[j]>dis[mark]+map[mark][j])
			dis[j]=dis[mark]+map[mark][j];
		}
	}
}
int main()
{
	int i,w,a,b,c;
	while(scanf("%d%d%d",&m,&n,&e)!=EOF)
	{
		memset(map,INF,sizeof(map));
		for(i=0;i<n;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			if(map[b][a]>c)
			map[b][a]=c;
		}
		dijkstra();
		scanf("%d",&w);
		int best=INF;
		for(i=0;i<w;i++)
		{
			scanf("%d",&a);
			best=best>dis[a]?dis[a]:best;
		}
		if(best==INF)
		printf("-1\n");
		else
		printf("%d\n",best);
	}
	return 0;
}

下面floyd代码虽然优化了，但是还是超时

#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int map[1010][1010],dis[1010],vis[1010];
int m,n,e,s[1010];
void floyd()
{
	int i,j,k;
	for(i=1;i<=m;i++)
	{
		for(j=1;j<=m;j++)
		{
			if(map[i][j]==INF)
			continue;
			for(k=1;k<=m;k++)
			{
				if(map[j][k]>map[j][i]+map[i][k])
				map[j][k]=map[j][i]+map[i][k];
			}
		}
	}
	
}
int main()
{
	int i,w,a,b,c;
	while(scanf("%d%d%d",&m,&n,&e)!=EOF)
	{
		memset(map,INF,sizeof(map));
		for(i=0;i<n;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			if(map[b][a]>c)
			map[b][a]=c;
		}
		floyd();
		scanf("%d",&w);
		int best=INF;
		for(i=0;i<w;i++)
		{
			scanf("%d",&a);
			best=best>map[e][a]?map[e][a]:best;
		}
		if(best==INF)
		printf("-1\n");
		else
		printf("%d\n",best);
	}
	return 0;
}