PAT 1072. Gas Station (30)

1072. Gas Station (30)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range.

Now given the map of the city and several candidate locations for the gas station, you are supposed to give the best recommendation. If there are more than one solution, output the one with the smallest average distance to all the houses. If such a solution is still not unique, output the one with the smallest index number.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive integers: N (<= 10³), the total number of houses; M (<= 10), the total number of the candidate locations for the gas stations; K (<= 10⁴), the number of roads connecting the houses and the gas stations; and D_S, the maximum service range of the gas station. It is hence assumed that all the houses are numbered from 1 to N, and all the candidate locations are numbered from G1 to GM.

Then K lines follow, each describes a road in the format
P1 P2 Dist
where P1 and P2 are the two ends of a road which can be either house numbers or gas station numbers, and Dist is the integer length of the road.

Output Specification:

For each test case, print in the first line the index number of the best location. In the next line, print the minimum and the average distances between the solution and all the houses. The numbers in a line must be separated by a space and be accurate up to 1 decimal place. If the solution does not exist, simply output “No Solution”.

Sample Input 1:

4 3 11 5
1 2 2
1 4 2
1 G1 4
1 G2 3
2 3 2
2 G2 1
3 4 2
3 G3 2
4 G1 3
G2 G1 1
G3 G2 2

Sample Output 1:

G1
2.0 3.3

Sample Input 2:

2 1 2 10
1 G1 9
2 G1 20

Sample Output 2:

No Solution

---

这道题是一个求最短路径的问题，用dijkstra可以解决。一开始最后一个点始终过不了，后来发现竟然是自己对输入的处理出错了（因为M<=10所以下意识的当成0~9，一位数来处理了，却忘了自己这次实际上是以1~10方式来计数）。。。汗。。。于是后来补了一个strtoint的函数用来处理数据，所有case就都过了。代码如下：

 

#include <iostream>
#include <algorithm>
#include <set>
#include <string>
#include <map>
#include <iomanip>
#include <cstring>
#include <climits>
using namespace std;
int dist[1012][1012];
int length[1012];
int visited[1012];
typedef struct info{
	bool flag;
	double ave;
	int min;
}info;
int N,M,K,Ds,i,size; //N:the total number of house; M:station; K:roads
map<int,info> result;
void init()
{
	memset(visited,0,sizeof(visited));
	for(int i=0;i<1011;i++)
		length[i]=INT_MAX;
}
void dijkstra(int station)
{
	init();
	length[station]=0;
	for(int i=1;i<=size;i++)
		if(dist[station][i]>0)
			length[i]=dist[station][i];
	int k=station;
	while(true)
	{
		visited[k]=1;
		for(int i=1;i<=size;i++)
		{
			if(!visited[i]&&dist[i][k]>0)
				if(dist[i][k]+length[k]<length[i])
					length[i]=dist[i][k]+length[k];
		}
		int minPos=k,min=INT_MAX;
		for(int i=1;i<=size;i++)
			if(!visited[i]&&length[i]<min)
			{
				min=length[i];
				minPos=i;
			}
		if(minPos==k)
			break;
		k=minPos;
	}
	int min=INT_MAX;
	double total=0;
	for(int i=1;i<=N;i++)
	{
		if(length[i]>Ds)
		{
			info temp;
			temp.flag=false;
			result[station-N]=temp;
			return;
		}
		else
		{
			if(length[i]<min)
				min=length[i];
			total+=length[i];
		}
	}
	info temp;
	temp.flag=true;
	temp.ave=total/N;
	temp.min=min;
	result[station-N]=temp;
}
int strtoint(string x)
{
	if(x[0]!='G')
		return atoi(x.c_str());
	else
	{
		string temp=x.substr(1);
		return atoi(temp.c_str())+N;
	}
}
int main(void)
{
	cin>>N>>M>>K>>Ds;
	size=M+N;
	memset(dist,0,sizeof(dist));
	while(K--)
	{
		string begin,end;
		int b,e,distance;
		cin>>begin>>end>>distance;
		b=strtoint(begin);
		e=strtoint(end);
		dist[b][e]=distance;
		dist[e][b]=distance;
	}
	for(int i=N+1;i<=size;i++)
		dijkstra(i);
	map<int,info>::iterator it;
	set<int> index;
	int min=0;
	double ave;
	for(it=result.begin();it!=result.end();it++)
	{
		if(it->second.flag==false)
			continue;
		if(it->second.min>min)
		{
			min=it->second.min;
			ave=it->second.ave;
			index.clear();
			index.insert(it->first);
		}
		else if(it->second.min==min)
		{
			if(it->second.ave<ave)
			{
				ave=it->second.ave;
				index.clear();
				index.insert(it->first);
			}
			else if(it->second.ave==ave)
				index.insert(it->first);
		}
	}
	if(index.size()==0)
	{
		cout<<"No Solution";
		return 0;
	}
	set<int>::iterator it2=index.begin();
	cout<<"G"<<*it2<<endl;
	cout<<fixed<<setprecision(1)<<result[*it2].min*1.0<<" "<<result[*it2].ave;
}