PAT (Advanced Level) Practise 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

题意：有n个居民点，m垃圾箱候选地点，垃圾箱的位置必须选在到所有居民点的最短距离最长的地方，同时还要保证每个居民点都在距离它一个不太远的范围内。如果解不唯一，则输出到所有居民点的平均距离最短的那个解。如果这样的解还是不唯一，则输出编号最小的地点。

解题思路：暴力枚举垃圾点的位置，求出到其他所有居民点的最短路，然后进行判断

#include <iostream>  
#include <cstdio>  
#include <string>  
#include <cstring>  
#include <algorithm>  
#include <cmath>  
#include <queue>  
#include <vector>  
#include <set>  
#include <stack>  
#include <map>  
#include <climits>  
  
using namespace std;  
  
#define LL long long  
const int INF=0x3f3f3f3f;  
  
int cnt,n,m,k,d;  
int s[10000],nt[100000],e[100000],l[100000];  
int dis[10000],visit[10000];  
  
struct node  
{  
    int id,dis;  
    friend bool operator <(node a,node b)  
    {  
        return a.dis>b.dis;  
    }  
}pre1,nt1;  
  
struct node1  
{  
    int id,midis;  
    double ave;  
}a[10000];  
  
int get(char ch[])  
{  
    int ans=0,len=strlen(ch),p=1;  
    if(ch[0]=='G')  
    {  
        for(int i=len-1; i>=1; i--)  
        {  
            ans+=(ch[i]-'0')*p;  
            p*=10;  
        }  
        ans+=n;  
    }  
    else  
    {  
        for(int i=len-1; i>=0; i--)  
        {  
            ans+=(ch[i]-'0')*p;  
            p*=10;  
        }  
    }  
    return ans;  
}  
  
void Dijkstra(int ss)  
{  
    pre1.id=ss,pre1.dis=0;  
    priority_queue<node>q;  
    memset(visit,0,sizeof visit);  
    memset(dis,INF,sizeof dis);  
    dis[ss]=0;  
    q.push(pre1);  
    while(!q.empty())  
    {  
        pre1=q.top();  
        q.pop();  
        visit[pre1.id]=1;  
        for(int i=s[pre1.id]; ~i; i=nt[i])  
        {  
            int ee=e[i];  
            if(visit[ee]) continue;  
            if(dis[ee]>dis[pre1.id]+l[i])  
            {  
                dis[ee]=dis[pre1.id]+l[i];  
                nt1.dis=dis[ee];  
                nt1.id=ee;  
                q.push(nt1);  
            }  
        }  
    }  
}  
  
bool cmp(node1 a,node1 b)  
{  
    if(a.midis!=b.midis) return a.midis>b.midis;  
    else if(a.ave!=b.ave) return a.ave<b.ave;  
    else return a.id<b.id;  
}  
  
int main()  
{  
    while(~scanf("%d %d %d %d",&n,&m,&k,&d))  
    {  
        cnt=1;  
        memset(s,-1,sizeof s);  
        memset(nt,-1,sizeof nt);  
        char s1[10],s2[10];  
        int ll;  
        for(int i=1; i<=k; i++)  
        {  
            scanf("%s %s %d",s1,s2,&ll);  
            int xx=get(s1),yy=get(s2);  
            nt[cnt]=s[xx],s[xx]=cnt,e[cnt]=yy,l[cnt++]=ll;  
            nt[cnt]=s[yy],s[yy]=cnt,e[cnt]=xx,l[cnt++]=ll;  
        }  
        int res=0;  
        for(int i=n+1;i<=n+m;i++)  
        {  
            Dijkstra(i);  
            double sum=0;  
            int flag=1,mi=INF;  
            for(int j=1;j<=n;j++)  
            {  
                if(dis[j]>d) {flag=0;break;}  
                mi=min(mi,dis[j]);  
                sum+=dis[j];  
            }  
            if(flag)  
            {  
                a[res].id=i;  
                a[res].midis=mi;  
                a[res++].ave=sum/n;  
            }  
        }  
        if(!res) {printf("No Solution\n");continue;}  
        sort(a,a+res,cmp);  
        printf("G%d\n",a[0].id-n);  
        printf("%.1lf %.1lf\n",(double)a[0].midis,a[0].ave);  
    }  
    return 0;  
}