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HDU 2680Choose the best route （SPFA）

2024-07-02 19:44:33   231人阅读

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

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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): 6285    Accepted Submission(s): 2067


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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注意：此题道路是单向的！

SPFA+反向寻找！

代码如下：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <malloc.h>
#include <ctype.h>
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define pi acos(-1.0)
#define INF 0xffffff
#define N 1047
#define M 20047
int n,m,s,pp,ww,k,Edgehead[M],dis[M];
struct
{
	int v,w,next;
}Edge[2*M];
bool vis[M];
void Addedge(int u,int v,int w)
{
	Edge[k].next = Edgehead[u];
	Edge[k].w = w;
	Edge[k].v = v;
	Edgehead[u] = k++;
}
void SPFA( int start)
{
	queue<int>Q;
	for(int i = 1 ; i <= N ; i++ )
		dis[i] = INF;
	dis[start] = 0;
	memset(vis,false,sizeof(vis));
	Q.push(start);
	while(!Q.empty())//直到队列为空
	{
		int u = Q.front();
		Q.pop();
		vis[u] = false;
		for(int i = Edgehead[u] ; i!=-1 ; i = Edge[i].next)//注意
		{
			int v = Edge[i].v;
			int w = Edge[i].w;
			if(dis[v] > dis[u] + w)
			{
				dis[v] = dis[u]+w;
				if( !vis[v] )//防止出现环，也就是进队列重复了  
				{
					Q.push(v);
					vis[v] = true;
				}
			}
		}
	}
}
int main()
{
	int u,v,w;
	while(~scanf("%d%d%d",&n,&m,&s))
	{
		k = 1;
		memset(Edgehead,-1,sizeof(Edgehead));
		for(int i = 1 ; i <= m ; i++ )
		{
			scanf("%d%d%d",&u,&v,&w);
		//	Addedge(u,v,w);
			Addedge(v,u,w);//因为是反向寻找的而且道路是单向的，所以此处也要方向
		}
		SPFA(s);
		scanf("%d",&pp);
		int min = INF; 
		for(i = 0 ; i < pp ;i++)
		{
			scanf("%d",&ww);
			if(min > dis[ww])
				min = dis[ww];
		}
		if(min == INF)
			min = -1;
		printf("%d\n",min);
	}
	return 0;
}

//有时正向不对，不防反向试试(生活亦如此)!

再附一份dijstra：

#include<stdio.h>
#define max 1001
#define INF 100000000

int N;
int v[max];
int dis[max];
int g[max][max];

void dijk()
{
	int i,j,mark,mindis;
	for(i=0;i<=N;i++)
	{
		v[i] = 0;
		dis[i] = INF;
	}
	dis[0] = 0;
	v[0]  = 1;
	for(i=0;i<=N;i++)
	{
		mindis = INF;
		mark = 0;
		for(j=0;j<=N;j++)
		if(!v[j] && dis[j] < mindis)
		{
			mindis = dis[j];
			mark = j;
		}
		v[mark] = j;
		for(j=0;j<=N;j++)
		if(!v[j] && dis[j] > dis[mark] + g[mark][j])
			dis[j] = dis[mark] + g[mark][j];
	}
}

int main()
{
	int W,M,Q,i,j,d,a,b;
	while(~scanf("%d%d%d",&N,&M,&Q))
	{
		for(i=0;i<max;i++)
		for(j=0;j<max;j++)
		{
			if(i==j)g[i][j] = 0;
			else
			g[i][j] = INF;
		}
		while(M--)
		{
			scanf("%d%d%d",&a,&b,&d);
			if(d < g[a][b])
			g[a][b] = d;
		}
		scanf("%d",&W);
		while(W--)
		{
			scanf("%d",&a);
			g[0][a] = 0;
		}
		dijk();
		if(dis[Q] == INF)
		printf("-1\n");
		else
		printf("%d\n",dis[Q]);
	}
}


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