PAT甲级1072 Gas Station (30 分)

1072 Gas Station (30 分)

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 (≤10​3​​), the total number of houses; M (≤10), the total number of the candidate locations for the gas stations; K (≤10​4​​), 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

      题目大意：

         给出n个房子与m个汽油站，k条边以及服务范围Ds。一个最佳汽油站取址应当是距离房子的最小距离中取得最大值的那个。如果答案不唯一则取平均距离最短的，如果还不唯一则取序号最小的。

 

思路：

         这是dijksta类型的题目，要求对dijkstra 算法比较熟悉。再者就是如果将Gx转化为整数，存储在图内的问题。算dijkstra类型题里较简单的一类，做好判定与处理即可。

 

参考代码：

 

#include
#include
using namespace std;
int n, m, k, ds;
const int inf = 0x7fffffff;
int e[1020][1020], dis[1020];
bool vis[1020];
void dijkstra(int v){
	fill(dis, dis + 1020, inf);
	fill(vis, vis + 1020, false);
	dis[v] = 0;
	for(int i = 1; i <= n + m; ++i){
		int u = -1, minv = inf;
		for(int j = 1; j <= n + m; ++j)
			if(minv > dis[j] && !vis[j])	minv = dis[j], u = j;
		if(u == -1)	return;
		vis[u] = true;
		for(int v = 1; v <= n + m; ++v )
			if(!vis[v] && e[u][v] && e[u][v] + dis[u] < dis[v])	dis[v] = dis[u] + e[u][v];
	}
}
int change(string &s){
	if(s[0] == 'G'){
		s.erase(s.begin());
		return n + stoi(s);
	} 
	return stoi(s);
}
int main(){
	scanf("%d%d%d%d", &n, &m, &k, &ds);
	for(int i = 0; i < k; ++i){
		string s1, s2;
		int temp, a, b;
		cin >> s1 >> s2 >> temp;
		a = change(s1), b = change(s2);
		 e[a][b] = e[b][a] = temp;
	}
	double mindis = 0, minavg = inf;
	int idx = -1;
	for(int i = 1; i <= m; ++i){
		dijkstra(i + n);
		sort(dis + 1, dis + n + 1);
		if(dis[n] > ds) continue;
		double avg = 0.0;
		for(int j = 1; j <= n; ++j)
			avg += 1.0 * dis[j] / n;
		if(dis[1] > mindis){
			mindis = dis[1];
			minavg = avg;
			idx = i;
		}else if(dis[1] == mindis && minavg > avg){
			minavg = avg;
			idx = i;
		}
	}
	if(idx == -1)	printf("No Solution");
	else	printf("G%d\n%.1f %.1f", idx, mindis, minavg);
	return 0;
}