【PAT】【Advanced Level】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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1072

https://www.nowcoder.com/pat/5/problem/4121

思路：

英语渣。。没怎看懂题意。。

题目的要求是：

针对多个解，首先保证到 最近点 的距离 最远。再保证平均值最小

DIJ+条件判断即可。

注意DIJ的细节（搜不到扩展点怎么办（退出），检查是否有不可达点）。。否则段错误。。

CODE：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdlib>
#include<cstdio>
#define N 1001
#define M 11


using namespace std;

int edge[N+M][N+M];

int dist[N+M];

bool flag[N+M];

int n,m,k,dm;

int dji(int s)
{
	for (int j=1;j<=m+n;j++)
	{
		dist[j]=edge[s][j];
		flag[j]=0;
	}
	flag[s]=1;
	for (int k=1;k<n+m;k++)
	{
		int fl=0;
		int minn=1000000000;
		int ind=-1;
		for(int i=1;i<=n+m;i++)
		{
			if (flag[i]==1) continue;
			if (dist[i]==-1) continue;
			if (dist[i]<minn)
			{
				minn=dist[i];
				ind=i;
			}
		}
		//cout<<ind<<endl;
		if (ind!=-1) flag[ind]=1;
		else
			break;
		for (int i=1;i<=n+m;i++)
		{
			if (flag[i]==1) continue;
			if (edge[ind][i]==-1) continue;
			if (dist[i]==-1||dist[i]>dist[ind]+edge[ind][i])
			{
				dist[i]=dist[ind]+edge[ind][i];
			}
			
		}
		//for (int i=1;i<=n+m;i++)	cout<<dist[i]<<" ";		cout<<endl;
	}
	int ma=-1;
	for (int i=1;i<=n;i++)
	{
		//cout<<dist[i]<<" ";
		if (i==s) continue;
		ma=max(ma,dist[i]);
		if (dist[i]==-1) return dm+1;
	}
	//cout<<endl;
	
	return ma;
}



int main()
{
	memset(edge,-1,sizeof(edge));
	cin>>n>>m>>k>>dm;
	for (int i=1;i<=n+m;i++){edge[i][i]=0;}
	for (int i=0;i<k;i++)
	{
		string aa,bb;
		int a,b,c;
		cin>>aa>>bb>>c;
		if (aa[0]=='G')
			a=atoi(aa.substr(1,aa.length()-1).c_str())+n;
		else
			a=atoi(aa.c_str());
		if (bb[0]=='G')
			b=atoi(bb.substr(1,bb.length()-1).c_str())+n;	
		else
			b=atoi(bb.c_str());	
		//cout<<edge[a][b]<<" "<<edge[b][a]<<endl;
		if (edge[a][b]==-1)
			edge[a][b]=edge[b][a]=c;
		else
			edge[a][b]=edge[b][a]=min(edge[a][b],c);
		//cout<<edge[a][b]<<" "<<edge[b][a]<<endl;
	}
	int inde=0;
	int mind=-1;
	int aver=1000000000;
	for (int i=1;i<=m;i++)
	{
		int re=dji(i+n);
		if (re<=dm)
		{
			int mi=1000000000;
			int ma=0;
			for (int j=1;j<=n;j++)
			{
				mi=min(mi,dist[j]);
				ma+=dist[j];
			}
			//cout<<mi<<endl;
			if (mi>mind)
			{
				inde=i;
				mind=mi;
				aver=ma;
			}
			else if (mi==mind)
			{
				if (ma<aver)
				{
					inde=i;
					mind=mi;
					aver=ma;	
				}
			}
		}
	}
	if (aver==1000000000)
	{
		printf("No Solution");
	}
	else
	{
		//cout<<mind<<" "<<aver;
		float re1=mind;
		float re2=(float)aver/(float)n;
		printf("G%d\n",inde);
		printf("%.1f %.1f",re1,re2);
	}
	return 0;
}