[TB-EXAM]小飛電梯-停一層(暫未考慮停N層)

電梯有N層，從底層到第i層的人有iFloorPassengerNum[i]個人，電梯只能停一次的話，求停那一次是的步行的層數最少。從下往上依次考察停某層的話，是Y-(NAbove-NFloor-NBottom)達到最小值， NAbove指到改層的人數， NBottom是該層停時到其以下層所有的人應該步行的層數，NAbove則是之上的。 最底層初始化這些數據，然後每考察一層時，其上每個人就少走一層，其下的每個人就多走一層，新到的一層爲到該層的人數。 但NAbove<NFloor+Nbottom時候即是過了V的谷底，之前的得出的幾位最小層數和該停的樓層。但可以停K層時，要考慮的問題增多，涉及到動態規劃，可能。主要是求得什麼樣的劃分是最好的。  

[程序]-

/******************************************************************* * * DESCRIPTION: Elevator Optimize Problem * * AUTHOR: Neesky * * DATE:2009-10-13 * *******************************************************************/ /*Marcor Definition*/ #define FLOORSNUM 9 /*include files*/ #include <iostream> using namespace std; /*Global Definition*/ int iFloorPassengerNum[FLOORSNUM]={0,4,4,5,3,2,2,1,3};/*每層乘客的數目-總共9層*/ /// <summary> /// Just For Show the Array /// </summary> /// <param name="Arr"></param> /// <param name="N"></param> /// <returns></returns> void OutputArray(int *Arr, int N) { for(int i=N-1;i>=0;--i) cout<<"/n "<<Arr[i]<<" ->No."<<i+1<<"F"; cout<<endl; return; } /// <summary> /// 求出在第I層停的所有情況下的WalkLayers,也是O(N)時間複雜度 /// </summary> /// <param name="_iFloorPassengerNum"></param> /// <param name="N"></param> /// <returns></returns> void StopAtOnlyOneFloor_AllFloorWalkLayers(int *_iFloorPassengerNum, int N) { cout<<"/n/nTotal Walks Numbers (All) : "; int NBottom=0,NFloor=_iFloorPassengerNum[0],NAbove=0,minWalkLayer=0; for(int i=1; i<N; ++i) { NAbove+=_iFloorPassengerNum[i]; minWalkLayer=minWalkLayer+(_iFloorPassengerNum[i]*i); } cout<<"/n "<<_iFloorPassengerNum[0]<<" ->No.1F -> "<<minWalkLayer<<"Fs"; for(int i=1; i<N; ++i) { NAbove=NAbove-_iFloorPassengerNum[i]; NBottom=NBottom+NFloor; NFloor=_iFloorPassengerNum[i]; minWalkLayer=minWalkLayer-NFloor-NAbove+NBottom; cout<<"/n "<<_iFloorPassengerNum[i]<<" ->No."<<i+1<<"F -> "<<minWalkLayer<<"Fs"; } cout<<endl; return; } /// <summary> /// If elevator only stop at one station, and for the walking fewest layers. O(n) /// Consider it from 1st Floor,NBottom:the bottom floors，NFloor:the floor NAbove:the above floor /// </summary> /// <param name="_iFloorPassengerNum"></param> /// <param name="N"></param> /// <returns></returns> void StopAtOnlyOneFloor_FewestWalkLayer(int *_iFloorPassengerNum,int N) { cout<<"/n**Floor Passengers Numbers: "; OutputArray(_iFloorPassengerNum,N); int NBottom=0, NFloor=_iFloorPassengerNum[0], NAbove=0,minWalkLayer=0; for(int i=1; i<N; ++i) //Consider from 1st Floor:i=0 { NAbove+=_iFloorPassengerNum[i]; minWalkLayer=minWalkLayer+(_iFloorPassengerNum[i]*i); } int j=1; for(;j<N;++j)/*Consider Stop At Above Floor: from j=1(The 2nd Floor)*/ { if(NAbove<NFloor+NBottom)break;/*Key，Passed the V's bottom，StopAtOnlyOneFloor_AllFloorWalkLayers不同的地方*/ NAbove=NAbove-_iFloorPassengerNum[j]; /*Up to the No.j+1 floor*/ NBottom=NBottom+NFloor; NFloor=_iFloorPassengerNum[j]; /*To get V's bottom*/ minWalkLayer=minWalkLayer-NAbove-NFloor+NBottom;/*Uppper Less 1 layer, Now and Bottom more 1 layer*/ } cout<<"**Best Stop is(FewestWalks): ->No."<<j<<"F, TotalWalks->"<<minWalkLayer<<"Fs "<<endl; return; } /// <summary> /// /// </summary> /// <param name="argc"></param> /// <param name="argv"></param> /// <returns></returns> int main (int argc, char *argv[]) { StopAtOnlyOneFloor_AllFloorWalkLayers(iFloorPassengerNum,FLOORSNUM); StopAtOnlyOneFloor_FewestWalkLayer(iFloorPassengerNum,FLOORSNUM); return(0); }

Total Walks Numbers (All)  : 

       0      ->No.1F     -> 92Fs

       4      ->No.2F     -> 68Fs

       4      ->No.3F     -> 52Fs

       5      ->No.4F     -> 44Fs

       3      ->No.5F     -> 46Fs

       2      ->No.6F     -> 54Fs

       2      ->No.7F     -> 66Fs

       1      ->No.8F     -> 82Fs

       3      ->No.9F     -> 100Fs


**Floor Passengers Numbers: 

       3      ->No.9F

       1      ->No.8F

       2      ->No.7F

       2      ->No.6F

       3      ->No.5F

       5      ->No.4F

       4      ->No.3F

       4      ->No.2F

       0      ->No.1F

**Best Stop is(FewestWalks): ->No.4F, TotalWalks->44Fs