1072. Gas Station (30)

最新推荐文章于 2024-10-22 21:50:18 发布

puppet_pyt

最新推荐文章于 2024-10-22 21:50:18 发布

阅读量289

点赞数

分类专栏： pat甲级文章标签： c++ pat dijkstra

本文链接：https://blog.csdn.net/puppet_pyt/article/details/71036713

版权

pat甲级专栏收录该内容

55 篇文章

订阅专栏

本文介绍了一个基于图论的算法，用于解决城市中加油站的最佳选址问题。该算法确保所有住宅都在加油站的服务范围内，并且尽可能地使最远住宅的距离最小化。在存在多个解决方案的情况下，选择平均距离最小且编号最小的加油站位置。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:

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

这道题目读懂就够费劲了，英语不好真的很伤，意思是首先你所有居民房都要在建设加油点服务范围Ds以内，如果有多种情况，则需要最近的一个房子距离比较远的方案，

如果也有相同则需要考虑所有房子总距离最小的方案，再者就是标号较小的建设点；

用的dijkstra，其实有时候挺纠结用dfs还是迪杰斯特拉，这道题目一开始用dfs尝试了下发现问题还挺多的，因为dfs只搜索一遍的话只是简单的遍历并不能得出最短路径，

还不是很熟练

#include <iostream>
#include <cctype>
#include <vector>
#include <utility>
#include <cstring>
#include <iterator>
#include <cstdio>
#include <iomanip>
#define MAX 1015
#define INF (0x6fffffff)
using namespace std;

struct Node{
    bool known;
    vector<pair<int, int> > v;
};

Node graph[MAX] = {false};
int dist[MAX] = {0};
int N, M, K, Ds;
void init()
{
    int w;
    int v1, v2;
    char vertex1[10], vertex2[10];
    for(int i=0; i<K; i++)
    {
        cin >> vertex1 >> vertex2 >> w;
        if(isdigit(vertex1[0]))
        {
            sscanf(vertex1, "%d", &v1);
        }
        else
        {
            sscanf(&vertex1[1], "%d", &v1);
            v1 += N;
        }
        if(isdigit(vertex2[0]))
        {
            sscanf(vertex2, "%d", &v2);
        }
        else
        {
            sscanf(&vertex2[1], "%d", &v2);
            v2 += N;
        }
        graph[v1].v.push_back(make_pair(v2, w));
        graph[v2].v.push_back(make_pair(v1, w));
    }
}
void dijkstra(int n)
{
    dist[n] = 0;
    int mindist, key;
    int temp = N+M;
    vector<pair<int, int> >::iterator it;
    while(temp--)
    {
        mindist = INF;
        for(int i=1; i<=N+M; i++)
        {
            if(dist[i] < mindist && !graph[i].known)
            {
                mindist = dist[i];
                key = i;
            }
        }
        graph[key].known = true;
        for(it=graph[key].v.begin(); it!=graph[key].v.end(); it++)
        {
            if(!graph[it->first].known)
            {
                if(it->second+mindist < dist[it->first])
                {
                    dist[it->first] = it->second+mindist;
                }
            }
        }
    }
}

int main()
{
    int k, ans, mind = -1;
    int finalsum = INF;
   // freopen("in.txt", "r", stdin);
    cin >> N >> M >> K >> Ds;
    init();
    for(int i=N+1; i<=N+M; i++)
    {
        for(int j=1; j<=N+M; j++)
        {
            dist[j] = INF;
            graph[j].known = false;
        }
        dijkstra(i);
        int sum = 0;
        int mindist = INF;
        for(k=1; k<=N; k++)
        {
            sum += dist[k];
            if(mindist > dist[k])
                mindist = dist[k];
            if(dist[k] > Ds)
            {
                break;
            }
        }
        if(k == N+1)
        {
            if(mindist > mind)
            {
                mind = mindist;
                ans = i;
                finalsum = sum;
            }
            else if(mindist == mind)
            {
                if(sum < finalsum)
                {
                    finalsum = sum;
                    ans = i;
                }
            }
        }
    }
    if(mind == -1)
    {
        cout << "No Solution" << endl;
    }
    else{
        cout << "G" << ans-N << endl;
        cout << fixed << setprecision(1) << mind*1.0 << " " << finalsum*1.0/N << endl;
    }
    return 0;

}