1072. Gas Station (30)

题目如下：

 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

题目的要求比较繁琐，一定要认真读题，下面对题意进行概括。

有N个住宅区、M个加气站，现在要求找出一个加气站，满足下面的要求：

①加气站到每个住宅区的距离都≤D

②加气站到每个住宅区的最短路径在满足①的所有加气站中最大。

③如果②出现了多个解，则找出加气站到所有住宅区距离之和最小的。

综上所述，这道题的关键就是找出每个加气站到所有住宅区的最短距离，也就是单源最短路径问题，使用Dijkstra算法对每个加气站都进行计算，然后分别存储，最后按照上面的要求筛选即可。

代码如下：

#include <iostream>
#include <vector>
#include <stdio.h>
#include <string>
#include <sstream>
#include <stdlib.h>
#include <memory.h>

using namespace std;

#define INF 99999999

struct Arc{
    double dist;
    int num;

    Arc(int n, double d){
        num = n;
        dist = d;
    }

};

int N,M,K,D;
bool* visited;
vector<vector<double> > minDists;
vector<vector<Arc> > graph;

int parseVertex(string ver){
    stringstream ss;
    ss << ver;
    int num;
    ss >> num;
    if(num != 0) return num;
    ss.clear();
    char c;
    ss >> c;
    ss >> num;
    return num + N;

}

void Dijkstra(int source){
    memset(visited,false,sizeof(bool)*(1 + N + M));
    minDists[source].resize(1 + N + M);
    for(int i = 1; i < minDists[source].size(); i++){
        minDists[source][i] = INF;
    }
    minDists[source][source] = 0;
    visited[source] = true;
    for(int i = 0; i < graph[source].size(); i++){
        Arc arc = graph[source][i];
        int num = arc.num;
        int dist = arc.dist;
        minDists[source][num] = dist;
    }
    double minDist = INF;
    int v;
    for(int k = 0; k < N + M; k++){
        minDist = INF;
        for(int i = 1; i < minDists[source].size(); i++){
            if(!visited[i] && minDists[source][i] < minDist){
                minDist = minDists[source][i];
                v = i;
            }
        }
        if(minDist == INF) break;
        visited[v] = true;
        for(int i = 0; i < graph[v].size(); i++){
            Arc arc = graph[v][i];
            int w = arc.num;
            double dist = arc.dist;
            if(visited[w]) continue;
            if(minDists[source][w] > minDists[source][v] + dist){
                minDists[source][w] = minDists[source][v] + dist;
            }
        }

    }

}

int main()
{
    cin >> N >> M >> K >> D;
    graph.resize(1 + N + M);
    visited = (bool*)malloc(1 + N + M);
    minDists.resize(1 + N + M);
    string source,dest;
    double dist;
    for(int i = 0; i < K; i++){
        cin >> source >> dest >> dist;
        int sourceInt = parseVertex(source);
        int destInt = parseVertex(dest);
        //printf("<%d %d %d>
",sourceInt,destInt,dist);
        graph[sourceInt].push_back(Arc(destInt,dist));
        graph[destInt].push_back(Arc(sourceInt,dist));
    }
    for(int i = 1; i <= M; i++){
        int gasNum = N + i;
        Dijkstra(gasNum);
    }
    int gasNum = -1;
    double gasSum;
    double maxMinDist = -1;
    double minDist = INF;
    double average;
    for(int i = N + 1; i < minDists.size(); i++){
        double sum = 0;
        minDist = INF;
        for(int j = 1; j <= N; j++){
            double dist = minDists[i][j];
            if(dist > D){
                sum = -1;
                break;
            }
            sum += dist;
            if(minDist > dist) minDist = dist;
        }
        if(sum == -1) continue;
        if(maxMinDist < minDist){
            gasNum = i - N;
            maxMinDist = minDist;
            gasSum = sum;
            average = sum / N;
        }else if(maxMinDist == minDist){
            if(sum < gasSum){
                gasNum = i - N;
                gasSum = sum;
                average = sum / N;
            }
        }
    }

    if(gasNum != -1){
        printf("G%d
%0.1f %0.1f
",gasNum,maxMinDist,average);
    }else{
        printf("No Solution
");
    }
    return 0;
}