[编程题]赶去公司
- 热度指数：4845 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 32M，其他语言64M
- 算法知识视频讲解
终于到周末啦！小易走在市区的街道上准备找朋友聚会，突然服务器发来警报,小易需要立即回公司修复这个紧急bug。假设市区是一个无限大的区域，每条街道假设坐标是(X，Y)，小易当前在(0，0)街道，办公室在(gx,gy)街道上。小易周围有多个出租车打车点，小易赶去办公室有两种选择，一种就是走路去公司，另外一种就是走到一个出租车打车点，然后从打车点的位置坐出租车去公司。每次移动到相邻的街道(横向或者纵向)走路将会花费walkTime时间，打车将花费taxiTime时间。小易需要尽快赶到公司去，现在小易想知道他最快需要花费多少时间去公司。
输入描述:
输入数据包括五行:
第一行为周围出租车打车点的个数n(1 ≤ n ≤ 50)
第二行为每个出租车打车点的横坐标tX[i] (-10000 ≤ tX[i] ≤ 10000)
第三行为每个出租车打车点的纵坐标tY[i] (-10000 ≤ tY[i] ≤ 10000)
第四行为办公室坐标gx,gy(-10000 ≤ gx,gy ≤ 10000),以空格分隔
第五行为走路时间walkTime(1 ≤ walkTime ≤ 1000)和taxiTime(1 ≤ taxiTime ≤ 1000),以空格分隔
输出描述:
输出一个整数表示，小易最快能赶到办公室的时间
示例1
输入
2 -2 -2 0 -2 -4 -2 15 3
输出
42
//
// main.cpp
// violent_enumeration
//
// Created by LU QI on 2017/11/06.
// Copyright © 2017 LU QI. All rights reserved.
//
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int spend(int x1,int y1,int x2,int y2,int Time){
int spend = abs(x1-x2)+abs(y1-y2);
spend *= Time;
return spend;
}
int main(int argc, const char * argv[]) {
int n,gx,gy,walkTime,taxiTime,t1,t2,t;
cin>>n;
int tx[n],ty[n];
for(int i=0;i<n;i++){
cin>>tx[i];
}
for(int i=0;i<n;i++){
cin>>ty[i];
}
cin>>gx>>gy>>walkTime>>taxiTime;
t = spend(tx[0], ty[0], 0, 0, walkTime)+spend(tx[0],ty[0],gx,gy,taxiTime);
for(int i=1;i<n;i++){
t1 = spend(tx[i],ty[i],0,0,walkTime);
t2 = spend(tx[i],ty[i],gx,gy,taxiTime);
if(t>t1+t2){
t = t1+t2;
}
}
int f = spend(0,0,gx,gy,walkTime);
t = t>f?f:t;
cout<<t<<endl;
return 0;
}
发表于 2017-11-08 13:22:07
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更多回答
第一眼看到题目以为是图的题。。。原来这么简单
package go.jacob.day913; import java.util.Scanner; public class Demo1 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int[] X = new int[n]; int[] Y = new int[n]; for (int i = 0; i < n; i++) { X[i] = sc.nextInt(); } for (int i = 0; i < n; i++) { Y[i] = sc.nextInt(); } int gx = sc.nextInt(), gy = sc.nextInt(); int walkTime = sc.nextInt(), taxiTime = sc.nextInt(); int min = walkTime * (Math.abs(gx) + Math.abs(gy)); for (int i = 0; i < n; i++) { int tmp = walkTime * (Math.abs(X[i]) + Math.abs(Y[i])) + taxiTime * (Math.abs(gx - X[i]) + Math.abs(gy - Y[i])); if (tmp < min) min = tmp; } System.out.println(min); sc.close(); } }
发表于 2017-09-13 09:58:44
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import java.util.*;
//该题目比较简单，针对两种不同的方式计算即可
//1. 完全走步
//2. 先走到停车点，之后坐车到达
public class Main{
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
//获得出租车打车点
int Tcount = scanner.nextInt();
int [] txArray = new int[Tcount];
int [] tyArray = new int[Tcount];
//初始化出租车打车点的x坐标
for(int i = 0;i < Tcount;i++){
txArray[i] = scanner.nextInt();
}
//初始化出租车打车点的Y坐标
for(int j = 0;j < Tcount;j++){
tyArray[j] = scanner.nextInt();
}
//得到目的地地址
int gx = scanner.nextInt();
int gy = scanner.nextInt();
//得到走路时间和打车时间
int walkTime = scanner.nextInt();
int taxiTime = scanner.nextInt();
//完全走路所需要的时间
int totalByWalk = (Math.abs(gx)+ Math.abs(gy)) * walkTime;
//保存打车所费时间的最小值
int totalByTaxi = Integer.MAX_VALUE;
//对于打车
for(int k = 0;k < Tcount;k++){
//走到停车点的时间
int firstTime = (Math.abs(txArray[k])+Math.abs(tyArray[k])) * walkTime;
//坐车到目的地时间
int secondTime = (Math.abs(txArray[k]- gx) + Math.abs(tyArray[k]- gy)) * taxiTime;
totalByTaxi = Math.min(totalByTaxi,firstTime + secondTime);
}
System.out.println(Math.min(totalByWalk,totalByTaxi));
}
}
发表于 2017-08-10 10:31:52
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import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int tx[] = new int[n];
int ty[] = new int[n];
for (int i = 0; i < n; i++) {
tx[i] = sc.nextInt();//打车点x坐标
}
for (int i = 0; i < n; i++) {
ty[i] = sc.nextInt();//打车点y坐标
}
int gx = sc.nextInt();//公司x坐标
int gy = sc.nextInt();//公司y坐标
int wt = sc.nextInt();// 走路速度
int dt = sc.nextInt();// 打车速度
int walkTime = (Math.abs(gx) + Math.abs(gy)) * wt;// 如果全部走路
int driveTime = Integer.MAX_VALUE;
for (int i = 0; i < n; i++) {// 如果打车
driveTime = Math.min(driveTime,
(Math.abs(ty[i]) + Math.abs(tx[i])) * wt + (Math.abs(ty[i]-gy)+Math.abs(tx[i]-gx)) * dt);
}
System.out.println(Math.min(driveTime, walkTime));
}
}
发表于 2017-04-04 23:00:06
回复(6)
#include<iostream>#include<vector>#include<map>#include<algorithm>#include<string>#include<cmath>using namespace std;intWal, Tex;intgetT(intx, inty, intrx, intry){// 计算走到车站的时间intwa = (abs(x)+abs(y))*Wal;// 计算车站到公司的时间intte = (abs(rx-x) + abs(ry-y))*Tex;// 返回总花费时间returnwa+te;}// 主函数intmain(){intn;inttx[51];intty[51];intcom[2];intwal, tex;cin >> n;inti=0;while(i!=n){cin>>tx[i++];}i=0;while(i!=n){cin>>ty[i++];}cin>>com[0]>>com[1];cin>>wal>>tex;Wal = wal;Tex = tex;// 直接步行的时间intwalktime = (abs(com[0]) + abs(com[1]))*wal;// 循环处理从每一个车站乘车去公司的时间intt = 2147483647;for(inti=0; i < n; i++){// 计算每一个车站乘车的总时间inttemp = getT(tx[i], ty[i], com[0], com[1]);// 判断是否是最小时间if(temp < t){t = temp;}}cout<<(t<walktime?t:walktime)<<endl;}
发表于 2017-03-25 17:31:22
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#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int tx[51];
int ty[52];
int n;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> tx[i];
}
for (int i = 0; i < n; i++)
{
cin >> ty[i];
}
int gx, gy;
cin >> gx >> gy;
int walkTime, taxiTime;
cin >> walkTime >> taxiTime;
int walk = (abs(gx) + abs(gy)) * walkTime;
int minTime = walk;
for (int i = 0; i < n; i++)
{
double temp = 0;
temp += (abs(tx[i]) + abs(ty[i])) * walkTime;
temp += (abs(tx[i] - gx) + abs(ty[i] - gy))*taxiTime;
if (temp < minTime)
minTime = temp;
}
cout << minTime;
}
#include<cmath>
using namespace std;
int main()
{
int tx[51];
int ty[52];
int n;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> tx[i];
}
for (int i = 0; i < n; i++)
{
cin >> ty[i];
}
int gx, gy;
cin >> gx >> gy;
int walkTime, taxiTime;
cin >> walkTime >> taxiTime;
int walk = (abs(gx) + abs(gy)) * walkTime;
int minTime = walk;
for (int i = 0; i < n; i++)
{
double temp = 0;
temp += (abs(tx[i]) + abs(ty[i])) * walkTime;
temp += (abs(tx[i] - gx) + abs(ty[i] - gy))*taxiTime;
if (temp < minTime)
minTime = temp;
}
cout << minTime;
}
发表于 2020-05-28 17:38:23
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import sys
n=int(sys.stdin.readline().strip())
x=list(map(int,sys.stdin.readline().strip().split()))
y=list(map(int,sys.stdin.readline().strip().split()))
targetx,targety=list(map(int,sys.stdin.readline().strip().split()))
walktime,taxitime=list(map(int,sys.stdin.readline().strip().split()))
result=walktime*(abs(targetx)+abs(targety))
for i in range(n):
t1=walktime*(abs(x[i])+abs(y[i]))
t2=taxitime*(abs(targetx-x[i])+abs(targety-y[i]))
t=t1+t2
if t<result:
result=t
print(result)
n=int(sys.stdin.readline().strip())
x=list(map(int,sys.stdin.readline().strip().split()))
y=list(map(int,sys.stdin.readline().strip().split()))
targetx,targety=list(map(int,sys.stdin.readline().strip().split()))
walktime,taxitime=list(map(int,sys.stdin.readline().strip().split()))
result=walktime*(abs(targetx)+abs(targety))
for i in range(n):
t1=walktime*(abs(x[i])+abs(y[i]))
t2=taxitime*(abs(targetx-x[i])+abs(targety-y[i]))
t=t1+t2
if t<result:
result=t
print(result)
发表于 2018-05-17 09:03:39
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### 分析
基础题，先算步行的时间，再枚举每个打车点算出打的时间，找到时间最短的即可。
### 我的答案
```cpp
#include<iostream>
#include<algorithm>
#define MAX 100000000
using namespace std;
int n;
int tx[51],ty[51];
int gx,gy;
int wt,dt;
int main(){
cin>>n;
int walkTime=MAX,driveTime=MAX;
for(int i=0;i<n;i++){
cin>>tx[i];
}
for(int i=0;i<n;i++){
cin>>ty[i];
}
cin>>gx>>gy>>wt>>dt;
walkTime=(abs(gx)+abs(gy))*wt;
for(int i=0;i<n;i++){
driveTime=min(driveTime,(abs(ty[i])+abs(tx[i]))*wt+(abs(ty[i]-gy)+abs(tx[i]-gx))*dt);
}
cout<<min(driveTime,walkTime)<<endl;
}
```
发表于 2017-03-31 13:08:22
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#include <iostream>
using namespace std;
struct Point {
int x;
int y;
} point;
int main()
{
int n;
cin >> n;
Point taxi[n];
for ( int i = 0; i < n; i++ )
cin >> taxi[i].x >> taxi[i].y;
Point company;
cin >> company.x >> company.y;
int walkTime, taxiTime;
cin >> walkTime >> taxiTime;
int res = ( abs(company.x) + abs(company.y) ) * walkTime;
int curWalk, curTaxi;
for ( int i = 0; i < n; i++ ) {
curWalk = ( abs(taxi[i].x) + abs(taxi[i].y) ) * walkTime;
curTaxi = ( abs(taxi[i].x - company.x) + abs(taxi[i].y - company.y) ) * taxiTime;
res = min(res, (curWalk + curTaxi) );
}
cout << res << endl;
return 0;
}
编辑于 2017-03-27 11:38:38
回复(3)
const timeCal = (posX, posY, offX, offY, speed) => {
const yTime = Math.abs(offY - posY) * speed
const xTime = Math.abs(offX - posX) * speed
return yTime + xTime
}
const main = (taxiNum, taxiX, taxiY, officeX, officeY, walkSpeed, taxiSpeed) => {
var time = timeCal(0, 0, officeX, officeY, walkSpeed);
for(let i = 0; i < taxiNum; i++){
let walkTime = timeCal(0, 0, taxiX[i], taxiY[i], walkSpeed)
let taxiTime = timeCal(taxiX[i], taxiY[i], officeX, officeY, taxiSpeed)
let totalTime = walkTime + taxiTime
time = (totalTime < time)?totalTime:time
}
return time
}
//except: 42
console.log(main(2, [-2, 0],[-2, -2], -4, -2, 15, 3))
发表于 2020-03-26 20:31:12
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#该题目比较简单，针对两种不同的方式计算即可
#完全走步
#先走到停车点，之后坐车到达，遍历所有停车点，不要觉得穷举就不能做
#有时候最直接的思路能解决就行
class solution():
def timeToCompany(self,n,tx,ty,gx,gy,walkTime,taxiTime):
n = int(n)
tx = [int(item) for item in tx]
ty = [int(item) for item in ty]
gx = int(gx)
gy = int(gy)
walkTime = int(walkTime)
taxiTime = int(taxiTime)
#完全走路需要的时间如下：
totalTimeWalk = (abs(gx)+abs(gy))*walkTime
#遍历所有打车点，求出最小的打车时间
totalTimeTaxi = float("inf")#保存打车所费时间的最小值
for i in range(n):
#走到停车点的时间
firstTime = (abs(tx[i])+abs(ty[i]))*walkTime
#坐车到公司的时间
secondTime = (abs(tx[i]-gx)+abs(ty[i]-gy))*taxiTime
totalTimeTaxi = min(totalTimeTaxi,firstTime+secondTime)
print(min(totalTimeTaxi,totalTimeWalk))
if __name__ == "__main__":
n = input()
tx = input().split(' ')
ty = input().split(' ')
gx,gy = input().split(' ')
walkTime,taxiTime = input().split(' ')
solution().timeToCompany(n,tx,ty,gx,gy,walkTime,taxiTime)
发表于 2019-08-24 18:17:55
回复(0)
#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<string>
#include<cmath>
#include<string>
using namespace std;
int main()
{
int n;
int tx[51],ty[51];
int gx,gy;
int wt,dt;
//cin>>n;
cin>>n;
//int i = 0;
for (int i = 0; i < n; i++) //横坐标
{
cin>>tx[i];
//i += 1;
}
//int j = 0;
for (int j = 0; j < n; j++) //纵坐标
{
cin>>ty[j];
//j += 1;
}
// for(int i=0;i<n;i++){
// cin>>tx[i];
//}
//for(int i=0;i<n;i++){
// cin>>ty[i];
//}
cin>>gx>>gy;
cin>>wt>>dt;
int OnlyWalkTime = (abs(gx) + abs(gy)) * wt;
for (int k = 0; k < n; k++)
{
int MixTime = (abs(gx - tx[k]) + abs(gy - ty[k])) * dt + (abs(tx[k]) + abs(ty[k])) * wt;
OnlyWalkTime = min(OnlyWalkTime, MixTime);
}
cout << OnlyWalkTime << endl;
return 0;
}
#include<vector>
#include<map>
#include<algorithm>
#include<string>
#include<cmath>
#include<string>
using namespace std;
int main()
{
int n;
int tx[51],ty[51];
int gx,gy;
int wt,dt;
//cin>>n;
cin>>n;
//int i = 0;
for (int i = 0; i < n; i++) //横坐标
{
cin>>tx[i];
//i += 1;
}
//int j = 0;
for (int j = 0; j < n; j++) //纵坐标
{
cin>>ty[j];
//j += 1;
}
// for(int i=0;i<n;i++){
// cin>>tx[i];
//}
//for(int i=0;i<n;i++){
// cin>>ty[i];
//}
cin>>gx>>gy;
cin>>wt>>dt;
int OnlyWalkTime = (abs(gx) + abs(gy)) * wt;
for (int k = 0; k < n; k++)
{
int MixTime = (abs(gx - tx[k]) + abs(gy - ty[k])) * dt + (abs(tx[k]) + abs(ty[k])) * wt;
OnlyWalkTime = min(OnlyWalkTime, MixTime);
}
cout << OnlyWalkTime << endl;
return 0;
}
发表于 2019-07-15 15:46:32
回复(0)
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
int main(){
int n,i,gx,gy,wt,tt,walk2stop,stop2company;
cin>>n;
int arr_x[n],arr_y[n];
for(i=0;i<n;++i) cin>>arr_x[i];
for(i=0;i<n;++i) cin>>arr_y[i];
cin>>gx>>gy>>wt>>tt;
int walktime=(abs(gx)+abs(gy))*wt;
int car_time=99999999;
for(i=0;i<n;++i){
walk2stop=(abs(arr_x[i])+abs(arr_y[i]))*wt;
stop2company=(abs(gx-arr_x[i])+abs(gy-arr_y[i]))*tt;
car_time=min(car_time,walk2stop+stop2company);
}
cout<<min(car_time,walktime);
return 0;
}
#include<math.h>
#include<algorithm>
using namespace std;
int main(){
int n,i,gx,gy,wt,tt,walk2stop,stop2company;
cin>>n;
int arr_x[n],arr_y[n];
for(i=0;i<n;++i) cin>>arr_x[i];
for(i=0;i<n;++i) cin>>arr_y[i];
cin>>gx>>gy>>wt>>tt;
int walktime=(abs(gx)+abs(gy))*wt;
int car_time=99999999;
for(i=0;i<n;++i){
walk2stop=(abs(arr_x[i])+abs(arr_y[i]))*wt;
stop2company=(abs(gx-arr_x[i])+abs(gy-arr_y[i]))*tt;
car_time=min(car_time,walk2stop+stop2company);
}
cout<<min(car_time,walktime);
return 0;
}
发表于 2018-12-19 22:00:45
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# 将步行，步行到打车点再打车的花费时间全部列举出来，找到最小值即可 import sys class Solution: def get_arrival_time(self, arr_x, arr_y, aim, walk_speed, taxi_speed): cost = self.get_walk_time(aim, walk_speed) index = 0 while index < len(arr_x): tmp_cost = self.get_taxi_time(aim, arr_x[index], arr_y[index], walk_speed, taxi_speed) cost = min([cost, tmp_cost]) index += 1 print(cost) def get_walk_time(self, aim, walk_speed): total_dis = abs(aim[0]) + abs(aim[1]) return total_dis * walk_speed def get_taxi_time(self, aim, x, y, walk_speed, taxi_speed): walk_time = self.get_walk_time((x, y), walk_speed) taxi_dis = abs(aim[0]-x) + abs(aim[1]-y) return walk_time + taxi_dis*taxi_speed if __name__ == '__main__': n = int(sys.stdin.readline()) arr_x = list(map(int, sys.stdin.readline().split())) arr_y = list(map(int, sys.stdin.readline().split())) aim = list(map(int, sys.stdin.readline().split())) w_speed, t_speed = map(int, sys.stdin.readline().split()) solution = Solution() solution.get_arrival_time(arr_x, arr_y, aim, w_speed, t_speed)
发表于 2018-07-11 10:26:54
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总是有10%的测试用例跑不过。谁能帮忙看看是哪里有问题吗？
#include <iostream>
using namespace std;
int main(int argc, const char * argv[]) {
int n;
cin>>n;
int tx[n], ty[n];
for (int i = 0; i < n; i++) {
cin>>tx[i];
}
for (int i = 0; i < n; i++) {
cin>>ty[i];
}
int gx, gy;
cin>>gx>>gy;
int walkT, taxiT;
cin>>walkT>>taxiT;
int totalWalkT = (abs(gx) + abs(gy)) * walkT; //完全走去的时间
int chosenTaxiX = 0, chosenTaxiY = 0, minLenToTaxi = 20000;
for (int i = 0; i < n; i++) {
int len = abs(tx[i]) + abs(ty[i]);
if (len < minLenToTaxi) {
minLenToTaxi = len;
chosenTaxiX = tx[i];
chosenTaxiY = ty[i];
}
}
int lenToG = abs(gx - chosenTaxiX) + abs(gy - chosenTaxiY);
int totalTaxiT = minLenToTaxi * walkT + lenToG * taxiT; //打车去的时间
cout<<(totalWalkT > totalTaxiT ? totalTaxiT : totalWalkT);
return 0;
}
发表于 2017-10-19 17:21:32
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package mem;
import java.util.*;
public class GoComp { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int taxNum = sc.nextInt(); int taxX[] = new int[taxNum]; int taxY[] = new int[taxNum]; for (int i = 0; i < taxNum; i++) { taxX[i] = sc.nextInt(); } for (int i = 0; i < taxNum; i++) { taxY[i] = sc.nextInt(); } int gx = sc.nextInt(); int gy = sc.nextInt(); int walkTime = sc.nextInt(); int taxTime = sc.nextInt(); int timeWalk = (Math.abs(gx) + Math.abs(gy)) * walkTime; int timeTax = timeWalk; for (int i = 0; i < taxNum; i++) { timeTax = Math.min(timeTax, ((Math.abs(taxX[i]) + Math.abs(taxY[i])) * walkTime + (Math.abs(gx - taxX[i]) + Math.abs(gx - taxX[i])) * taxTime)); } System.out.println(timeTax); }
}
发表于 2017-09-17 20:21:18
回复(0)
importjava.util.*;
publicclassMain{
publicstaticvoidmain(String[] args)
{
Scanner sc=newScanner(System.in);
intn=sc.nextInt();
intx[]=newint[n];
for(inti=0;i<n;i++){
x[i]=sc.nextInt();
}
inty[]=newint[n];
for(inti=0;i<n;i++){
y[i]=sc.nextInt();
}
intgx=sc.nextInt();
intgy=sc.nextInt();
intwt=sc.nextInt();
inttt=sc.nextInt();
inttime=0;
if(wt<=tt){
time=(Math.abs(gx)+Math.abs(gy))*wt;
System.out.println(time);
}
else{
time=(Math.abs(x[0])+Math.abs(y[0]))*wt+(Math.abs(x[0]-gx)+Math.abs(y[0]-gy))*tt;
for(inti=0;i<n;i++){
time=Math.min(time,(Math.abs(x[i])+Math.abs(y[i]))*wt+(Math.abs(x[i]-gx)+Math.abs(y[i]-gy))*tt);
}
time=Math.min(time,(Math.abs(gx)+Math.abs(gy))*wt);
System.out.println(time);
}
}
}
发表于 2017-08-14 14:01:56
回复(0)
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int main(){
int taxi_num = 0;
cin >> taxi_num;
vector<int> tx;
vector<int> ty;
int i = 0,j = 0;
for(int i = 0;i < taxi_num;i++){
int x;
cin >> x;
tx.push_back(x);
}
for(int i = 0;i < taxi_num;i++){
int y;
cin >> y;
ty.push_back(y);
}
int fx,fy;
cin>>fx>>fy;
int wt,tt;
cin>>wt>>tt;
int min_time = 0x7FFFFFFF;
int sum = 0,nx,ny,tl,time2;
for(int i = 0; i < taxi_num;i++){
sum = abs(tx[i])+ abs(ty[i]);
nx = tx[i];
ny = ty[i];
tl = abs(fx-nx) + abs(fy -ny);
time2 = sum * wt + tl * tt; //走到最近的出租车点，坐车
if(min_time > time2){
min_time = time2;
}
}
int time1 = wt * (abs(fx) + abs(fy));//只是步行
if(min_time < time1)
cout << min_time << endl;
else
cout << time1 << endl;
return 0;
}
发表于 2017-08-12 12:09:40
回复(0)
问题信息
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2022-07-13 20:56:04
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