PAT 1072. Gas Station (30)

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


此题是一个常规题，主要用了dijkstra算法，其他的就是一些细节的处理了：

#include<iostream>

#include <map>

#include <vector>

#include <memory.h>

#include <algorithm>

#include <numeric>

#include <climits>

using namespace std;

const int N=1001;

const int M=11;

bool visited[M+N];

double Distance[M+N][M+N];

double minDis[M+N];
map<int,vector<int> > adjlist;
struct Gas

{

    int index;

    double minDis;

    double averageDis;

    bool operator<(const Gas& rhs) const

    {

        if(minDis!=rhs.minDis)

            return minDis>rhs.minDis;

        if(averageDis!=rhs.averageDis)

            return averageDis<rhs.averageDis;

        return index<rhs.index;

    }

}GasStations[M];
int str2int(char p[],int n)

{

    int indicator=0;

    if(p[0]==‘G‘)

        indicator=1;

    int index=indicator==0?0:n;

    index+=atoi(p+indicator);

    return index;

};
void reset()

{

    memset(visited,false,sizeof(bool)*(N+M));

    for(int i=0;i<N+M;++i)

        minDis[i]=-1.0;

}
void dijkstra(int source,int n)

{

    reset();

    visited[source]=true;

    minDis[source]=0;

    int newP=source;

    for(int i=1;i<n;++i)

    {

        size_t size=adjlist[newP].size();

        for(int j=0;j<size;++j)

        {

            int child=adjlist[newP][j];

            if(visited[child])

                continue;

            if(minDis[child]<0||minDis[child]>minDis[newP]+Distance[newP][child])

                minDis[child]=minDis[newP]+Distance[newP][child];

        }

        int min=INT_MAX;

        for(int k=1;k<=n;++k)

        {

            if(visited[k]||minDis[k]<0)

                continue;

            if(minDis[k]<min)

            {

                min=minDis[k];

                newP=k;

            }

        }

        visited[newP]=true;

    }

}
int main(int argc, char *argv[])

{

    int n,m,k,Ds;

    scanf("%d %d %d %d",&n,&m,&k,&Ds);

    int i=0;

    char p1[5],p2[5];

    int idx1,idx2;

    for(;i<k;++i)

    {

        scanf("%s %s",p1,p2);

        idx1=str2int(p1,n);

        idx2=str2int(p2,n);

        scanf("%lf",&Distance[idx1][idx2]);

        Distance[idx2][idx1]=Distance[idx1][idx2];

        adjlist[idx1].push_back(idx2);

        adjlist[idx2].push_back(idx1);

    }

    int candidates=0;

    for(i=0;i<m;++i)

    {

        int source=n+i+1;

        dijkstra(source,n+m);

        int j=1;

        for(;j<=n;++j)

            if(minDis[j]>Ds)

                break;

        if(j==n+1)

        {

            GasStations[candidates].index=i+1;

            GasStations[candidates].minDis=*min_element(minDis+1,minDis+n+1);

            GasStations[candidates].averageDis=accumulate(minDis+1,minDis+n+1,0);

            ++candidates;

        }

    }

    if(0==candidates)

    {

        printf("No Solution\n");

        return 0;

    }

    sort(GasStations,GasStations+candidates);

    printf("G%d\n%.1f %.1f\n",GasStations[0].index,GasStations[0].minDis,GasStations[0].averageDis/n);

    return 0;

}

PAT 1072. Gas Station (30)