A no of changes added from different forks

C++/abstract.tex

@@ -9,6 +9,8 @@ \subsubsection{内容简介}
 
 全书的代码，使用“纯C + STL”的风格。本书中的代码规范，跟在公司中的工程规范略有不同，为了使代码短（方便迅速实现）:
 
+
+test file
 \begindot
 \item 所有代码都是单一文件。这是因为一般OJ网站，提交代码的时候只有一个文本框，如果还是
 按照标准做法，比如分为头文件.h和源代码.cpp，无法在网站上提交；

C++/src/maxNetProfit.cpp

@@ -0,0 +1,89 @@
+#include<iostream>
+using namespace std;
+/*
+Author: Jiankai Sun
+Date: 2014.10.15
+*/
+/*
+N jobs are to be scheduled for processing on one machine. Job i, 1 ≤ i ≤ N,
+needs ti units of processing time. If job i is finished by time T, where T is a given
+deadline, then a profit pi is earned; otherwise, a penalty qi
+is imposed. (Both pi and qi are positive integers.) We want to select a subset S of jobs such that
+(i) ∑i∈S ti ≤ T, and (ii) f(S) = ∑i∈S pi −∑i /∈S qi is maximum.
+Show how to find such a set of jobs using dynamic programming.
+*/
+int const N=3;
+int const T=6;
+//profit
+int p[N]={12,3,4};
+//penalty
+int q[N]={1,4,2};
+//time consuming
+int t[N]={1,4,3};
+//net profit table
+int f[N+1][T+1];
+//pointer to index which job should be scheduled
+int P[N+1][T+1];
+
+void findMaxNetProfit(){
+	//non recursive implementation
+	for(int i=0;i<N;i++){
+		for(int j=0;j<T;j++){
+			//k is the job index; from 1 to N
+			int k=i+1;
+			// j is the time unit; form 1 to T
+			int l=j+1;
+			if(t[i]>l){
+				f[k][l]=f[k-1][l]-q[i];
+			}
+			else{
+				f[k][l]=max(f[k-1][l]-q[i],f[k-1][l-t[i]]+p[i]);
+				if(f[k][l]==f[k-1][l-t[i]]+p[i]){
+					P[k][l]=1;
+				}
+			}
+		}
+	}
+	cout<<"net profit is "<<f[N][T]<<endl;
+}
+int findMaxNetProfit_rec(int job_n,int deadline){
+	//recursive implementation 
+	if(job_n==0){
+		//boundary condition
+		return 0;
+	}
+	if(t[job_n-1]>deadline){
+		return findMaxNetProfit_rec(job_n-1,deadline)-q[job_n-1];
+	}
+	else{
+		int a=findMaxNetProfit_rec(job_n-1,deadline)-q[job_n-1];
+		int b=findMaxNetProfit_rec(job_n-1,deadline-t[job_n-1])+p[job_n-1];
+		if(b>a)
+			P[job_n][deadline]=1;
+		return max(a,b);
+	}
+
+}
+void printJobSet(){
+	int l=T;
+	for(int i=N;i>=1;i--){
+			if(P[i][l]==1){
+				cout<<"job "<<i<<" is in job set"<<endl;
+				l=l-t[i-1];
+			}
+	}
+}
+int main(){
+
+	for(int i=0;i<i+1;i++)
+		f[0][0]=0;
+	for(int i=0;i<N;i++){
+		for(int j=0;j<T;j++){
+			P[i][j]=0;
+		}
+	}
+	//findMaxNetProfit();
+	cout<<findMaxNetProfit_rec(N,T)<<endl;
+	printJobSet();
+	return 0;
+}

C++/src/max_selected_intervals.cpp

@@ -0,0 +1,76 @@
+/*
+Author: Jiankai Sun
+Date: 2014.10.21
+*/
+#include<iostream>
+using namespace std;
+//find the largest j which makes f[j]<=s[i]
+int find_opt(double [], double [],int);
+//return the maximum sum of the selected intervals
+//non recurssive version
+double max_select_nonrec(double [],double [],int);
+//recurssive version
+double max_select_rec(double [],double [],int);
+//void print selected intervals
+void print_interval(int *,int);
+
+int find_opt(double s[],double f[],int i){
+	for(int j=i-1;j>=0;j--){
+		if(f[j]<=s[i]){
+			return j;
+		}
+	}
+	return 0;
+}
+
+double max_select_nonrec(double s[], double f[],int n){
+	double * T=new double[n];
+	int * P=new int[n];
+	for(int i=0;i<n;i++){
+		T[i]=0;
+	}
+	for(int i=1;i<n;i++){
+		int j=find_opt(s,f,i);
+		double temp=T[j]+f[i]-s[i];
+		if(temp>=T[i-1]){
+			T[i]=temp;
+			P[i]=j;
+		}
+		else{
+			T[i]=T[i-1];
+			P[i]=-1;
+		}
+
+	}
+	//print 
+	print_interval(P,n);
+	return T[n-1];
+}
+void print_interval(int * P, int n){
+	int i=n-1;
+	while(P[i]!=0){
+		if(P[i]>0){
+			cout<<i<<" ";
+			i=P[i];
+		}
+	}
+	//print the last one
+	cout<<i<<endl;
+}
+
+double max_select_rec(double s[],double f[], int i){
+
+	if(i<=0)
+		return 0;
+	else
+		return max(max_select_rec(s,f,find_opt(s,f,i))+f[i]-s[i],max_select_rec(s,f,i-1));
+}
+
+int main(){
+	int n=7;
+	double s[]={0,1,5,6,7,15,1};
+	double f[]={0,4,10,12,14,20,21};
+	cout<<max_select_rec(s,f,n-1)<<endl;
+	cout<<max_select_nonrec(s,f,n)<<endl;
+	return 0;
+}

C++/src/min_operation_dp.cpp

@@ -0,0 +1,54 @@
+#include<iostream>
+using namespace std;
+/*
+Author: Jiankai Sun
+Date: 2014/10/14
+*/
+/*
+Let A=a1a2...am and B=b1b2...bn be two strings of characers. we
+want to transform A into B using following operations:
+delete a character
+and a character
+change a character
+write a dynamic programming algorithm that finds the minimum number
+of operations needed to transform A into B
+*/
+int min_ope(char * c1,const int m, char *c2,const int n){
+
+	int F[m+1][n+1];//define operation matrix
+
+	//boundary condition
+	for(int i=0;i<=m;i++){
+		F[i][0]=i;		
+	}
+	for(int i=0;i<=n;i++){
+		F[0][i]=i;
+	}
+	for(int i=0;i<m;i++){
+		for(int j=0;j<n;j++){
+			//the index of F starts from [1,1] to store c1[0] and c2[0]
+			int k=i+1;
+			int l=j+1;
+			if(c1[i]==c2[j])
+				//equal, don't need to change
+				F[k][l]=F[k-1][l-1];
+			else{
+				// add, delete or change operation
+				F[k][l]=min(min(F[k-1][l-1]+1,F[k-1][l]+1),F[k][l-1]+1);
+			}
+
+		}
+	}
+	return F[m][n];
+}
+
+int main(){
+int const m=2;
+int const n=1;
+char c1[m]={'a','c'};
+char c2[n]={'b'};
+
+cout<<min_ope(c1,m,c2,n)<<endl;
+
+return 0;
+}