【PAT】【Advanced Level】1072. Gas Station (30)
来源:互联网 发布:php连不上mysql数据库 编辑:程序博客网 时间:2024/06/04 23:27
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 51 2 21 4 21 G1 41 G2 32 3 22 G2 13 4 23 G3 24 G1 3G2 G1 1G3 G2 2Sample Output 1:
G12.0 3.3Sample Input 2:
2 1 2 101 G1 92 G1 20Sample Output 2:
No Solution
https://www.patest.cn/contests/pat-a-practise/1072
https://www.nowcoder.com/pat/5/problem/4121
思路:
英语渣。。没怎看懂题意。。
题目的要求是:
针对多个解,首先保证到 最近点 的距离 最远。再保证平均值最小
DIJ+条件判断即可。
注意DIJ的细节(搜不到扩展点怎么办(退出),检查是否有不可达点)。。否则段错误。。
CODE:
#include<iostream>#include<cstring>#include<string>#include<cstdlib>#include<cstdio>#define N 1001#define M 11using namespace std;int edge[N+M][N+M];int dist[N+M];bool flag[N+M];int n,m,k,dm;int dji(int s){for (int j=1;j<=m+n;j++){dist[j]=edge[s][j];flag[j]=0;}flag[s]=1;for (int k=1;k<n+m;k++){int fl=0;int minn=1000000000;int ind=-1;for(int i=1;i<=n+m;i++){if (flag[i]==1) continue;if (dist[i]==-1) continue;if (dist[i]<minn){minn=dist[i];ind=i;}}//cout<<ind<<endl;if (ind!=-1) flag[ind]=1;elsebreak;for (int i=1;i<=n+m;i++){if (flag[i]==1) continue;if (edge[ind][i]==-1) continue;if (dist[i]==-1||dist[i]>dist[ind]+edge[ind][i]){dist[i]=dist[ind]+edge[ind][i];}}//for (int i=1;i<=n+m;i++)cout<<dist[i]<<" ";cout<<endl;}int ma=-1;for (int i=1;i<=n;i++){//cout<<dist[i]<<" ";if (i==s) continue;ma=max(ma,dist[i]);if (dist[i]==-1) return dm+1;}//cout<<endl;return ma;}int main(){memset(edge,-1,sizeof(edge));cin>>n>>m>>k>>dm;for (int i=1;i<=n+m;i++){edge[i][i]=0;}for (int i=0;i<k;i++){string aa,bb;int a,b,c;cin>>aa>>bb>>c;if (aa[0]=='G')a=atoi(aa.substr(1,aa.length()-1).c_str())+n;elsea=atoi(aa.c_str());if (bb[0]=='G')b=atoi(bb.substr(1,bb.length()-1).c_str())+n;elseb=atoi(bb.c_str());//cout<<edge[a][b]<<" "<<edge[b][a]<<endl;if (edge[a][b]==-1)edge[a][b]=edge[b][a]=c;elseedge[a][b]=edge[b][a]=min(edge[a][b],c);//cout<<edge[a][b]<<" "<<edge[b][a]<<endl;}int inde=0;int mind=-1;int aver=1000000000;for (int i=1;i<=m;i++){int re=dji(i+n);if (re<=dm){int mi=1000000000;int ma=0;for (int j=1;j<=n;j++){mi=min(mi,dist[j]);ma+=dist[j];}//cout<<mi<<endl;if (mi>mind){inde=i;mind=mi;aver=ma;}else if (mi==mind){if (ma<aver){inde=i;mind=mi;aver=ma;}}}}if (aver==1000000000){printf("No Solution");}else{//cout<<mind<<" "<<aver;float re1=mind;float re2=(float)aver/(float)n;printf("G%d\n",inde);printf("%.1f %.1f",re1,re2);}return 0;}
- 【PAT】【Advanced Level】1072. Gas Station (30)
- PAT (Advanced Level) Practise 1072 Gas Station (30)
- PAT (Advanced Level) Practise 1072 Gas Station (30)
- Pat(Advanced Level)Practice--1072(Gas Station)
- PAT (Advanced Level) 1072. Gas Station (30) Dijkstra最短路径+剪枝
- 1072. Gas Station (30)【最短路dijkstra】——PAT (Advanced Level) Practise
- PAT 1072. Gas Station (30)
- PAT 1072. Gas Station (30)
- 【PAT】1072. Gas Station (30)
- PAT 1072. Gas Station (30)
- PAT 1072. Gas Station
- PAT 1072. Gas Station
- PAT 1072. Gas Station
- 【PAT】1072. Gas Station
- pat 1072. Gas Station
- PAT A 1072. Gas Station (30)
- PAT甲级练习1072. Gas Station (30)
- 1072. Gas Station (30) PAT 甲级
- Iframe跨域访问
- HDU6144 Arithmetic of Bomb
- 关于流通系统的流失特性的研究
- BZOJ 4800 [Ceoi2015]Ice Hockey World Championship ……
- Java-mysql-批量插入
- 【PAT】【Advanced Level】1072. Gas Station (30)
- js闭包(closure)
- Redis为什么是单线程
- Eclipse自动补全功能轻松设置
- MyBatis 详细介绍
- 考究Hadoop中split的计算方法
- 【PAT】Set Similarity&&Student List for Course
- 解决`向github提交代码是老要输入用户名密码`的问题
- J2EE进阶之JDBC数据库连接池 十九