hdu 2680 Choose the best route

最新推荐文章于 2021-07-20 11:08:36 发布

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最新推荐文章于 2021-07-20 11:08:36 发布

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分类专栏： ACM算法_最短路文章标签：最短路

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本文详细解析了一个关于寻找从多个起点到单一终点的最短路径问题，并提供了一段实现该算法的C++代码示例。文章通过具体示例介绍了如何初始化距离矩阵、如何更新最短路径，最终求得最少时间成本。

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Choose the best route

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16181 Accepted Submission(s): 5271

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

有多个起点，只有一个终点，问最短路径，还是单向图。

设起源点0，0到多个起点的距离为0，就把问题转化的简单了。

hdu 2066这个最全，http://blog.youkuaiyun.com/zhang__liuchen/article/details/78580323，这个会了的话，这个题就简单了。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int inf=0x3f3f3f3f;
int cost[1010][1010];
int dp[1010];
int n,m,s;
void Dijkstra()
{
	int vis[1010]={0};
	int i,j,pos,mi;
	for(i=0;i<=n;i++)
		dp[i]=cost[0][i];
	dp[0]=0;
	vis[0]=1;
	for(i=0;i<=n;i++)
	{
		mi=inf;
		for(j=0;j<=n;j++)
		{
			if(!vis[j]&&mi>dp[j])
			{
				pos=j;
				mi=dp[j];
			}
		}
		if(mi==inf)
			break;
		vis[pos]=1;
		for(j=0;j<=n;j++)
		{
			if(dp[j]>dp[pos]+cost[pos][j]&&!vis[j])
				dp[j]=dp[pos]+cost[pos][j];
		}
	}
}
int main()
{
	while(scanf("%d%d%d",&n,&m,&s)!=EOF)
	{
		for(int i=0;i<=n;i++)
		{
			for(int j=0;j<=n;j++)
				cost[i][j]=dp[i]=inf;
			cost[i][i]=0;
		}
		int p,q,t;
		while(m--)
		{
			scanf("%d%d%d",&p,&q,&t);
			if(cost[p][q]>t)//竟然必须加这个
				cost[p][q]=t;
		}
		int w,cnt;
		scanf("%d",&w);
		while(w--)
		{
			scanf("%d",&cnt);
			cost[0][cnt]=0;
		}
		Dijkstra();
		if(dp[s]==inf)
			printf("-1\n");
		else
			printf("%d\n",dp[s]);
	}
	return 0;
}

参考博客

http://blog.youkuaiyun.com/niushuai666/article/details/6794343