1072. Gas Station (30)

1.题目要求一个加油站，能够到达所有的屋子，并且这个加油站到屋子的最小距离尽可能地大（实际上也应该，不然汽油挥发会影响人身安全啊啊啊）

2.根据汽油站index排序时，不能使用string，要转换成int排序，避免出现G1<G123<G2<G23这样的情况

3.通过dijkstra求每个加油站到各个屋子的距离，即dijkstra外加一个循环遍历。然后统计所有的最短路径和平均路径，进行排序

4.最小距离一样的，根据平均距离来排序，平均距离仍然一样的，根据加油站的id来排序

5.为了提高空间利用率，把加油站和屋子都统一成一个结构体，这个结构体数据的0~housesum用来表示屋子，housesum+1~housesum+1+gassum表示加油站



AC代码：

//#include<string>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include <iomanip>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;
/*
2 1 2 20
1 G1 9
2 G1 20

1 1 1 20
1 G1 9

4 3 11 5
1 2 2
1 4 2
1 G1 5
1 G2 3
2 3 2
2 G2 1
3 4 2
3 G3 2
4 G1 3
G2 G1 1
G3 G2 2

*/
struct Node{
vector<pair<int,int>> list;//邻居
bool visited;
bool sured;
long long cost;
Node() :list(0), visited(false), sured(false), cost(INT_MAX){};
};
string num2string(int a)
{
if (a == 0) return "0";
string ans = "";
while (a != 0)
{
char c = a % 10 + '0';
ans = c + ans;
a /= 10;
}
return ans;

}
struct resultNode{
string gas;//不能够用string排序，避免G1,G123,G2,G23这样的情况
int id;
double min;
double avg;
double totalCost;
resultNode(string s,int i, double m, double a) :gas(s), id(i), min(m), avg(a), totalCost(0){};
resultNode() :gas(""), id(0), min(0), avg(0), totalCost(0){};
};
bool cmp(const resultNode&a, const resultNode&b)
{
if (a.min > b.min)
return true;
else if (a.min == b.min && a.totalCost < b.totalCost)
return true;
else if (a.min == b.min && a.totalCost == b.totalCost && a.id < b.id)
return true;
else return false;

}
int main(void)
{
int houseSum, gasSum, roadSum, serviceRange;
cin >> houseSum>> gasSum>>roadSum>>serviceRange;
vector<Node> place( houseSum+gasSum+1);//1~houseSum表示房子，houseSum+1~houseSum+gasSum表示加油站
map<int,bool> gasStationNum;

for (int i = 0; i < roadSum; i++)
{
int a = 0;
char tmp[5];
scanf("%s", tmp);
if (tmp[0] == 'G')//加油站
{
a = 0;//1~houseSum表示房子，houseSum+1~houseSum+gasSum表示加油站
for (int j = 1; j < 5 && tmp[j] != 0; j++)
a = a * 10 + tmp[j] - '0';
a += houseSum;//1~houseSum表示房子，houseSum+1~houseSum+gasSum表示加油站
}
else
{
a = 0;
for (int j = 0; j < 5 && tmp[j] != 0; j++)
a = a * 10 + tmp[j] - '0';
}
int b = 0;
scanf("%s", tmp);
if (tmp[0] == 'G')//加油站
{
b = 0;//1~houseSum表示房子，houseSum+1~houseSum+gasSum表示加油站
for (int j = 1; j < 5 && tmp[j] != 0; j++)
b = b * 10 + tmp[j] - '0';
b += houseSum;//1~houseSum表示房子，houseSum+1~houseSum+gasSum表示加油站
}
else
{
b = 0;
for (int j = 0; j < 5 && tmp[j] != 0; j++)
b = b * 10 + tmp[j] - '0';
}

int cost;
scanf("%d", &cost);
place[a].list.push_back({ b, cost });
place[b].list.push_back({ a, cost });
}

double minCost = INT_MAX;
double totalCost = INT_MAX;
vector<resultNode> ans(0);
for (int nowGas = houseSum + 1; nowGas < houseSum + gasSum + 1; nowGas++)
{
bool canChoose = true;//记录是否能够选为加油站，后面服务范围会使用
long long thisMinCost = INT_MAX;
long long thisTotalCost = 0;
vector<Node> v = place;//避免统计每个加油站时，数据已经存在，所以每次计算加油站，都新建一个节点vector

//使用dijkstra算法统计单元最短路径
v[nowGas].visited = true;
v[nowGas].cost = 0;
while (1)
{
int p = -1;
for (int i = 0; i < v.size(); i++)
{
if (p == -1 && v[i].visited&&!v[i].sured)
p = i;
else if (p != -1 && v[i].visited && !v[i].sured && v[i].cost < v[p].cost)
p = i;
}
if (p == -1) break;
v[p].sured = true;
if (p <= houseSum && v[p].cost > serviceRange)
{//如果该点是房子，但是在服务距离之外，则false
canChoose = false;
break;
}
for (int i = 0; i < v[p].list.size(); i++)
{
int q = v[p].list[i].first;
if (!v[q].sured && v[q].cost > v[p].cost + v[p].list[i].second)
{
v[q].visited = true;//标记为已经探望
v[q].cost = v[p].cost + v[p].list[i].second;//更新最短路径
}
}
}
if (!canChoose) continue;//有些房子服务范围之外，不能选为gas
for (int i = 1; i <= houseSum; i++)
{//只统计房子部分，计算最短路径和总耗费
thisMinCost = min(thisMinCost, v[i].cost);
thisTotalCost += v[i].cost;
if (!v[i].sured)
{//有些房子没有到达，不能选为gas
canChoose = false;
break;
}
}
if (!canChoose) continue;//有些房子没有到达，不能选为gas

//存储结果
minCost = thisMinCost;
totalCost = thisTotalCost;
string gasID = "G";
gasID += num2string(nowGas - houseSum);
double avg = thisTotalCost*1.0 / houseSum;
ans.push_back(resultNode(gasID, nowGas - houseSum, thisMinCost, avg));
ans.back().totalCost = thisTotalCost;

}
if (ans.size() == 0)
printf("No Solution\n");
else
{
sort(ans.begin(), ans.end(), cmp);
cout << ans[0].gas << endl;
printf("%.1lf %.1lf\n", ans[0].min, ans[0].avg);
}

return 0;
}