Elevator dispatching--c++--Software engineering

Source: Internet
Author: User

First, design ideas

(1) Save the number of floors the passengers are going to.

(2) Assuming that Yi is the passenger to climb the number of floors of the sum, yi=n1*| (N1-ni) |+n2*| (N2-ni) |+. +n18*| (N18-ni) |

(3) Compare the values of Y1 to Y18 to find the smallest.

(4) Output results

Second, the source code

#include "iostream.h" #include "stdlib.h" #define MAXSIZE 15void input (int &num,int flour[])//Enter the number of people travelling by elevator and    Floor {cout<< "************** Elevator scheduling applet ***************" <<endl;    cout<< "Please enter the number of people who take the elevator (limited to 15 persons):";    cin>>num;cout<<endl; if (num>15) {cout<< "Input error!        "<<endl;    Exit (0);    } cout<< "Please enter the number of layers each person is going to (up to 18 storeys):";        for (int i=0;i<num;i++) {cin>>flour[i]; if (flour[i]>18) {cout<< "Input error!            "<<endl;        Exit (0);    }}}void getsmallestflour (int num,int flour[],int &small)//calculates the floor {int Sum,temp=small with the lowest number of stairs to climb;    int highest,lowest;    HIGHEST=LOWEST=FLOUR[0]; for (int i=0;i<num;i++)//Find the highest floor in the floor to go {if (flour[i]>highest) {highest=f        Lour[i];        } else//find the lowest floor {lowest=flour[i]; }} for (i=lowest;i<=highest;i++)//from the lowest floorThe layer begins to traverse until the highest floor {sum=0; for (int j=0;j<num;j++) {if ((flour[j]-i) <0)//Calculates the sum of the number of floors of the stair ladder {su            M=SUM+I-FLOUR[J];            } else {sum=sum+flour[j]-i;        }}cout<<endl;        cout<< "When the elevator stops at the" <<i<< "floor, the passenger climbs the stairs with the sum of" <<sum<<endl;cout<<endl;            if (sum<temp) {small=i;            Record and the smallest floor number temp=sum;    Record the smallest and}}}int main (int argc, char* argv[]) {int num,aver,flour[maxsize];    Input (Num,flour);    aver=0;    for (int i=0;i<num;i++) {aver+=flour[i];    } getsmallestflour (Num,flour,aver); cout<< "The elevator will stop at the <<aver<<" floor.    "<<endl;cout<<endl; return 0;}


Iii. Results of operation

Iv. Experience

Although I think this elevator algorithm is not human-friendly, but it is the most powerful solution to the problem of clogging. The main problem is that logic starts a little messy.

Elevator dispatching--c++--Software engineering

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