PAT甲級1072 Gas Station (30 分)題解

$\quad$這個題要求每個加油站到所有居住地的最短距離，最長距離和平均距離。其中最長距離用於判斷該加油站位置是否符合要求，最短距離和平均距離需要輸出並保留一位小數。我用1到N表示居民居住地，N+1到N+M+1表示加油站，這樣計算出某個加油站到其他地點的距離，保存在數組中，再進行處理。
$\quad$在所有位置合法的加油站中，以最大化該加油站到居住地的最短距離爲第一標準，最小化加油站到居住地的平均距離爲第二標準輸出即可，若沒有合法的結果則輸出"No Solution"。

#include <iostream>
#include <vector>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int maxn = 1011;
vector<pair<int, int> > E[maxn];
int vis[maxn], dis[maxn];
int N, M, K, Ds;
struct Station
{
	int id, minDis;
	float avg;
	bool legal;
}station[maxn];
bool cmp(const Station &s1, const Station &s2)
{
	if(s1.minDis!=s2.minDis) return s1.minDis>s2.minDis;
	else if(s1.avg!=s2.avg) return s1.avg<s2.avg;
	else return s1.id<s2.id;
}
void Dijistra(int s)
{
	memset(vis, 0, sizeof(vis));
	fill(dis, dis+maxn, 0x3f3f3f3f);
	dis[s] = 0;
	priority_queue<pair<int, int> > q;
    q.push({0, s});
	while(!q.empty())
	{
		int u = q.top().second;
		q.pop();
		if(vis[u]==1) continue;
		vis[u] = 1;
		for (int i = 0; i < E[u].size(); ++i)
		{
		 	int v = E[u][i].first, w = E[u][i].second;
		 	if(dis[v]>dis[u]+w)
		 	{
		 		dis[v] = dis[u]+w;
		 		if(vis[v]==0) q.push({-dis[v], v});
		 	}
		}
	}
    station[s].id=s, station[s].minDis=0x3f3f3f3f, station[s].legal=true;
    int sum = 0;
    for (int i = 1; i <= N; ++i)
    {
    	if(dis[i]>Ds) station[s].legal = false;
    	sum += dis[i];
    	station[s].minDis = min(station[s].minDis, dis[i]);
    }
    station[s].avg = sum*1.0/N;
}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> N >> M >> K >> Ds;
    while(K--)
    {
    	string a, b; int w;
    	cin >> a >> b >> w;
    	int u, v;
    	if(a[0]=='G'){
    		sscanf(a.c_str(), "G%d", &u);
    		u += N;
    	}
    	else sscanf(a.c_str(), "%d", &u);
    	if(b[0]=='G'){
    		sscanf(b.c_str(), "G%d", &v);
    		v += N;
    	}
    	else sscanf(b.c_str(), "%d", &v);
    	E[u].push_back({v, w});
    	E[v].push_back({u, w});
    }
    for (int i = N+1; i <= N+M; ++i) Dijistra(i);
    sort(station+N+1, station+N+M+1, cmp);
	bool flag = false;
	for(int i = N+1; i <= N+M; i++)
	{
		if(station[i].legal)
		{
			cout << "G" << station[i].id-N << endl;
			printf("%.1f %.1f\n", 1.0*station[i].minDis, station[i].avg);
			flag = true;
			break;
		}
	}
	if(!flag) cout << "No Solution" << endl;
	return 0;
}