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 DS, 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 此题不难,只是考察了最简单的Dijkstra算法,读清楚题意即可,代码如下:
#include<iostream>
#include<string>
#include<iomanip>
#include"stdio.h"
using namespace std; int N,M,K,Ds;
int **G; bool *collected;
int *dist; void Dijkstra(int);
int GetMinDist(); int main()
{
scanf("%d%d%d%d",&N,&M,&K,&Ds); //初始化G
G = new int* [N+M];
dist = new int [N+M];
collected = new bool [N+M];
for (int i=;i<N+M;i++)
{
G[i] = new int [N+M];
collected[i] = false;
dist[i] = ;
}
for (int i=;i<N+M;i++)
for (int j=;j<N+M;j++)
G[i][j] = ; int iP1,iP2;
string P1,P2;
int Dist; //开始输入道路信息
for (int i=;i<K;i++)
{
cin >> P1 >> P2 >> Dist;
if (P1[] == 'G')
{
P1.erase(P1.begin());
iP1 = atoi(P1.c_str())+N-;
}
else
iP1 = atoi(P1.c_str())-; if (P2[] == 'G')
{
P2.erase(P2.begin());
iP2 = atoi(P2.c_str())+N-;
}
else
iP2 = atoi(P2.c_str())-; G[iP1][iP2] = Dist;
G[iP2][iP1] = Dist;
} //开始Dijkstra
int mind=;
double averd=.;
bool flag_all = true;
int minG=-,mindd=-;
double averdd=-.; for (int i=N;i<N+M;i++) //对每一个G进行Dijkstra
{
Dijkstra(i);
//对dist[]进行扫描,找出最小和平均值,或者没有
for (int j=;j<N;j++)
{
averd += dist[j];
if (mind > dist[j])
mind = dist[j];
if (dist[j] > Ds)
{
flag_all = false;
break;
}
}
if (flag_all)
{
averd = averd / double(N);
if (mindd < mind)
{
mindd = mind;
minG=i;
averdd = averd;
}
else if (mindd == mind)
{
if (averdd > averd)
{
minG=i;
averdd = averd;
}
}
} //复原各变量
for (int j=;j<N+M;j++)
{
collected[j] = false;
dist[j] = ;
}
mind=;
averd=.;
flag_all = true;
} if (minG == -)
cout << "No Solution" << endl;
else
{
cout << 'G' << minG-N+ << endl;
cout << fixed << setprecision() << double(mindd) << ' ' << fixed << setprecision() << averdd << endl;
} return ;
} void Dijkstra(int S)
{
//初始化源节点
collected[S]=true;
dist[S]=; //初始化与S相邻的节点
for (int i=;i<N+M;i++)
{
if (G[S][i])
{
dist[i]=G[S][i];
}
} //开始贪心算法
int V;
while ()
{
V = GetMinDist();
if (V == -)
break;
collected[V]=true; for (int i=;i<N+M;i++)
{
if (!collected[i] && G[V][i] && dist[i] > dist[V]+G[V][i])
dist[i] = dist[V]+G[V][i];
}
}
} int GetMinDist()
{
int min_d = ,min_d_i=-; for (int i=;i<N+M;i++)
if (!collected[i] && dist[i] < min_d)
{
min_d = dist[i];
min_d_i = i;
}
return min_d_i;
}
PAT 1072. Gas Station (30)的更多相关文章
- 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 sta ...
- 1072. Gas Station (30)【最短路dijkstra】——PAT (Advanced Level) Practise
题目信息 1072. Gas Station (30) 时间限制200 ms 内存限制65536 kB 代码长度限制16000 B A gas station has to be built at s ...
- PAT 甲级 1072 Gas Station (30 分)(dijstra)
1072 Gas Station (30 分) A gas station has to be built at such a location that the minimum distance ...
- PAT 1072 Gas Station[图论][难]
1072 Gas Station (30)(30 分) A gas station has to be built at such a location that the minimum distan ...
- PAT Advanced 1072 Gas Station (30) [Dijkstra算法]
题目 A gas station has to be built at such a location that the minimum distance between the station an ...
- 1072. Gas Station (30)
先要求出各个加油站 最短的 与任意一房屋之间的 距离D,再在这些加油站中选出最长的D的加油站 ,该加油站 为 最优选项 (坑爹啊!).如果相同D相同 则 选离各个房屋平均距离小的,如果还是 相同,则 ...
- 1072. Gas Station (30) 多源最短路
A gas station has to be built at such a location that the minimum distance between the station and a ...
- 1072 Gas Station (30)(30 分)
A gas station has to be built at such a location that the minimum distance between the station and a ...
- PAT 1072. Gas Station
A gas station has to be built at such a location that the minimum distance between the station and a ...
随机推荐
- 使用cocos2d-x c++ Android静态库
在用cocos2d-x做Android开发时,每次clean后都会需要再次编译coco2d-x的库,十分耗时. 这里给出一个直接使用静态库而不用每次都编译源码的方法: 1\ 首先找到一个cocos2d ...
- OpenBSD内核之引导MBR
MBR的介绍网上很多,没错,就那个最后以0x55AA结尾的512字节的引导块,OpenBSD提供了引导MBR实现:OpenBSD在x86上的引导过程为MBR --> PBR --> boo ...
- gulp教程之gulp-less
简介: 使用gulp-less插件将less文件编译成css,当有less文件发生改变自动编译less,并保证less语法错误或出现异常时能正常工作并提示错误信息. 1.安装nodejs/全局安装gu ...
- CentOS_7.2安装Redis_3.0
一.安装依赖包和开发工具: yum install vim vim-enhanced wget zip unzip telnet ntsysv compat* apr* nasm* gcc gcc* ...
- 一个在线jpg png转ICO的网站
网站地址: https://lvwenhan.com/convertico/
- UISegmentControl
@import url(http://i.cnblogs.com/Load.ashx?type=style&file=SyntaxHighlighter.css);@import url(/c ...
- .NET的面向对象
一.继承 1.C#中的继承规则 继承是可传递的 派生类是对基类的扩展 构造函数和析构函数不能被继承 派生类可以覆盖已继承的成员 派生类只能从一个类中继承,可以通过接口来实现多重继承 类可以定义虚属性. ...
- 解决OS X系统连接VPN后无法访问内网资源的问题
该问题是第一次使用OS X系统连接VPN遇到的问题,现象是连接VPN成功,但无法访问公司的内网资源. 主要原因还是VPN设置上的问题,在系统偏好设置中打开VPN连接,里面有个高级设置,如图: 点击高级 ...
- js 10秒倒计时 功能
请等待<span id=</span>秒 <script type="text/javascript"> function run(){ var s ...
- monads-are-elephants(转)
介绍monads有点像互联网时代的家庭手工业.我想 “为什么要反对传统?”,但这篇文章将以Scala对待monads的方式来描述. 有个古老的寓言,讲述了几个瞎子第一次摸到大象.一个抱着大象的腿说:“ ...