pat 甲级 1072. Gas Station (30)

1072. Gas Station (30)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题意：最短路，找出一个建立加油站的合适地方。现在有几个备选的地方。按照如下规则筛选：
1：加油站与所有住宅区的的最小距离越大越好。
2：加油站与所有住宅区的平均距离越小越好。
3：挑选编号数值最小的加油站。
AC 代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<queue>
#include<map>
using namespace std;
#define INF 0x3f3f3f
#define N_MAX 1000+20
typedef long long ll;
int n,m,k,d_max;
struct edge {int to, cost;edge() {}edge(int to ,int cost):to(to),cost(cost) {}
};
vector<edge>G[N_MAX];
struct P {int first, second;P() {}P(int first,int second):first(first),second(second) {}bool operator < (const P&b)const {return first > b.first;} 
};
int d[N_MAX];
int V;
void dijkstra(int s) {priority_queue<P>que;fill(d, d + V, INF);d[s] = 0;que.push(P(0,s));while (!que.empty()) {P p = que.top(); que.pop();int v = p.second;if (d[v] < p.first)continue;for (int i = 0; i < G[v].size();i++) {edge e = G[v][i];if (d[e.to]>d[v]+e.cost) {d[e.to] = d[v] + e.cost;que.push(P(d[e.to], e.to));}}}
}int translation(string s) {if (s[0] == 'G') {if (s.size() == 3)return m+n;//m最大为10，唯一的三位数else return s[1] - '0'+n;}else {return atoi(s.c_str());}
}
string recover(int id) {string s="G";s += '0' + id-n;return s;
}int main() {while (cin>>n>>m>>k>>d_max) {V = n + m+1;for (int i = 0; i < k;i++) {string from, to; int cost;cin >> from >> to >> cost;G[translation(from)].push_back(edge(translation(to), cost));G[translation(to)].push_back(edge(translation(from), cost));}double max_mindist = -1, max_avedist = -1; int id;for (int i = 1; i <= m;i++) {//对于每一个stationbool flag = 1;//判断当前情况是否可以int pos = n + i;dijkstra(pos);double tmp_ave = 0;int tmp_min = INF;for (int j = 1; j <= n; j++) {if (d[j] > d_max) {flag = 0;break;}tmp_min = min(d[j], tmp_min);tmp_ave += d[j];}if (!flag)continue;tmp_ave /= (double)n;if (tmp_min > max_mindist) {max_mindist=tmp_min;max_avedist = tmp_ave;id = pos;}else if (tmp_min == max_mindist&&tmp_ave < max_avedist) {max_avedist=tmp_ave;id = pos;}else if (tmp_min == max_mindist&&tmp_ave == max_avedist&& id>pos) {id = pos;}}if (max_mindist == -1)puts("No Solution");else {cout << recover(id) << endl;printf("%.1f %.1f\n",max_mindist,max_avedist);}}return 0;
}