PAT_A1072#Gas Station

Source:

PAT A1072 Gas Station (30 分)

Description:

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 (≤), the total number of houses; M (≤), the total number of the candidate locations for the gas stations; K (≤), 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

Keys:

• 最短路径

Attention:

• 加油站可以作为中间结点

Code:

 /*
 Data: 2019-06-18 17:04:24
 Problem: PAT_A1072#Gas Station
 AC: 43:23

 题目大意：
 加油站选取的最佳位置，在服务范围内，离居民区的最短距离尽可能的远
 如果最佳位置不唯一，挑选距离居民区平均距离最近，且编号最小的位置
 输入：
 第一行给出，房子数N，候选加油站数M，总路径数K，最大服务范围Ds
 接下来K行，v1,v2,dist
 输出：
 最佳位置编号
 最小距离，平均距离（一位小数）
 没有则No Solution
 */
 #include<cstdio>
 #include<string>
 #include<iostream>
 #include<vector>
 #include<algorithm>
 using namespace std;
 const int M=1e3+,INF=1e9;
 int grap[M][M],vis[M],d[M];
 int n,m;

 int Str(string s)
 {
     if(s[] == 'G')
     {
         s = s.substr();
         return n+atoi(s.c_str());
     }
     else
         return atoi(s.c_str());
 }

 void Dijskra(int s)
 {
     fill(d,d+M,INF);
     fill(vis,vis+M,);
     d[s]=;
     for(int i=; i<=n+m; i++)
     {
         int u=-, Min=INF;
         for(int j=; j<=n+m; j++)
         {
             if(vis[j]== && d[j]<Min)
             {
                 Min = d[j];
                 u = j;
             }
         }
         if(u==-)    return;
         vis[u]=;
         for(int v=; v<=n+m; v++)
             if(vis[v]== && grap[u][v]!=INF)
                 if(d[u]+grap[u][v] < d[v])
                     d[v] = d[u]+grap[u][v];
     }
 }

 int main()
 {
 #ifdef  ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     fill(grap[],grap[]+M*M,INF);

     int k,v1,v2,Ds;
     scanf("%d%d%d%d", &n,&m,&k,&Ds);
     for(int i=; i<k; i++)
     {
         string index;
         cin >> index;
         v1 = Str(index);
         cin >> index;
         v2 = Str(index);
         scanf("%d", &grap[v1][v2]);
         grap[v2][v1]=grap[v1][v2];
     }
     int maxSum=INF,minDist=,id=-;
     for(int i=n+; i<=n+m; i++)
     {
         Dijskra(i);
         int sum=,dist=INF;
         for(int j=; j<=n; j++)
         {
             if(d[j] > Ds)
             {
                 dist=-;
                 break;
             }
             sum += d[j];
             if(d[j] < dist)
                 dist=d[j];
         }
         if(dist==-)
             continue;
         if(dist > minDist){
             minDist = dist;
             maxSum = sum;
             id = i;
         }
         else if(dist==minDist && sum<maxSum){
             maxSum=sum;
             id = i;
         }
     }
     if(id == -)
         printf("No Solution");
     else
         printf("G%d\n%.1f %.1f", id-n, (1.0)*minDist,(1.0)*maxSum/n);

     return ;
 }