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算法导论第十五章练习题15.1-1c++代码实现

2012-09-09 
算法导论第十五章习题15.1-1c++代码实现利用递归方式以站号递增的方式输出装配站:本代码采用了两种递归方

算法导论第十五章习题15.1-1c++代码实现

利用递归方式以站号递增的方式输出装配站:

本代码采用了两种递归方式,第一种用了两个函数来实现递

归,第二种用了一个函数实现递归,第二种思路相对麻烦,它是将前一种方法的两个函数封装到一个函数中。

具体代码如下:

//动态规划,工厂装配线#include<iostream>using namespace std;//e1,e2是两条装配生产线的输入耗费x1,x2是输出耗费,a1[i]是第一条生产线的第i个站耗时,t1,t2记录了两条生产线转换时耗时//f1[i]为记录第一条生产线上第i个站生产耗时最少时间,L1[i]记录了第一条生产线上第i个站生产耗时最少时,前一站是那条生产线上的//length为每条生产线上的站数void FastWay(int e1,int e2,int *a1,int *a2,int x1,int x2,int *f1,int *f2,int *L1,int *L2,int &L,int *t1,int *t2,int length){f1[1]=e1+a1[1];f2[1]=e2+a2[1];int i;for(i=2;i<length;i++){if(f1[i-1]+a1[i]<f2[i-1]+a1[i]+t2[i-1]){f1[i]=f1[i-1]+a1[i];L1[i]=1;}else{f1[i]=f2[i-1]+a1[i]+t2[i-1];L1[i]=2;}if(f2[i-1]+a2[i]<f1[i-1]+a2[i]+t1[i-1]){f2[i]=f2[i-1]+a2[i];L2[i]=2;}else{f2[i]=f1[i-1]+a2[i]+t1[i-1];L2[i]=1;}}//利用f1[length]来存储所用最短时间if(f1[length-1]+x1<f2[length-1]+x2){f1[length]=f1[length-1]+x1;L=1;}else{f1[length]=f2[length-1]+x2;L=2;}}//输出最短时间以及所需要的路线void Printanswser(int *L1,int *L2,int L,int length,int *f1){int i=L;cout<<"the fast time is : "<<f1[length+1]<<endl;cout<<"the station : "<<length<<" is on line: "<<i<<endl;int j;for(j=length;j>=2;j--){if(i==1){i=L1[j];}else{i=L2[j];}cout<<"the station : "<<j-1<<" is on line: "<<i<<endl;}}//递归实现结果顺序打印,通过两个函数来实现void PrintAnswer(int *L1,int *L2,int L,int length){int i;if(L==1){i=L1[length+1];}else{i=L2[length+1];}if(length>=2){PrintAnswer(L1,L2,i,length-1);}if(i==1){cout<<"station is : "<<length<<" line is : "<<L1[length+1]<<endl;}else{cout<<"station is : "<<length<<" line is : "<<L2[length+1]<<endl;}}void Print(int *L1,int *L2,int L,int length,int *f1){cout<<"the fast time is : "<<f1[length+1]<<endl;PrintAnswer(L1,L2,L,length-1);int i=L;cout<<"station is : "<<length<<" line is : "<<L<<endl;}//递归实现结果顺序打印,通过一个函数来实现void PrintAnswerInOneFun(int *L1,int *L2,int L,int len,int length,int *f1){int i;if(L==1){i=L1[len+1];}else{i=L2[len+1];}if(len>=2){PrintAnswerInOneFun(L1,L2,i,len-1,length,f1);}if(i==1){cout<<"station is : "<<len<<" line is : "<<L1[len+1]<<endl;}else{cout<<"station is : "<<len<<" line is : "<<L2[len+1]<<endl;}if(len==length-1){i=L;cout<<"station is : "<<length<<" line is : "<<L<<endl;cout<<"the fast time is : "<<f1[length+1]<<endl;}}int main(){int e1=2,e2=4,x1=3,x2=2;int a1[8]={0,7,9,3,4,8,4};int a2[8]={0,8,5,6,4,5,7};int t1[6]={0,2,3,1,3,4};int t2[6]={0,2,1,2,2,1};int f1[8],f2[7],L1[7],L2[7],L;FastWay(e1,e2,a1,a2,x1,x2,f1,f2,L1,L2,L,t1,t2,7);Printanswser(L1,L2,L,6,f1);cout<<endl;Print(L1,L2,L,6,f1);cout<<endl;PrintAnswerInOneFun(L1,L2,L,5,6,f1);return 0;}


 

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