C++与Java分别解决活动选择问题和带权活动选择问题

活动安排问题

活动选择问题代码实现

带权活动选择问题

带权活动选择问题代码实现

贪心算法总是作出在当前看来最好的选择。也就是说贪心算法并不从整体最优考虑，它所作出的选择只是在某种意义上的局部最优选择。

活动安排问题

问题描述：设有n个活动的集合E = {1,2,…,n}，其中每个活动都要求使用同一资源，如演讲会场等，而在同一时间内只有一个活动能使用这一资源。每个活i都有一个要求使用该资源的起始时间si和一个结束时间fi,且si < fi 。如果选择了活动i，则它在半开时间区间[si, fi)内占用资源。若区间[si, fi)与区间[sj, fj)不相交，则称活动i与活动j是相容的。也就是说，当si >= fj或sj >= fi时，活动i与活动j相容。

活动选择问题代码实现

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std ;
struct ActivityTime
{
public:
    ActivityTime (int nStart, int nEnd) 
        : m_nStart (nStart), m_nEnd (nEnd) 
    { }
    ActivityTime ()
        : m_nStart (0), m_nEnd (0)
    { }
    friend 
    bool operator < (const ActivityTime& lth, const ActivityTime& rth) 
    {
        return lth.m_nEnd < lth.m_nEnd ;
    }
public:
    int m_nStart ;
    int m_nEnd ;
} ;
class ActivityArrange 
{
public:
    ActivityArrange (const vector<ActivityTime>& vTimeList) 
    {
        m_vTimeList = vTimeList ;
        m_nCount = vTimeList.size () ;
        m_bvSelectFlag.resize (m_nCount, false) ;
    }
    // 活动安排
    void greedySelector () 
    {
        __sortTime () ;
        // 第一个活动一定入内
        m_bvSelectFlag[0] = true ;    
        int j = 0 ;
        for (int i = 1; i < m_nCount ; ++ i) {
            if (m_vTimeList[i].m_nStart > m_vTimeList[j].m_nEnd) {
                m_bvSelectFlag[i] = true ;
                j = i ;
            }
        }
        copy (m_bvSelectFlag.begin(), m_bvSelectFlag.end() ,ostream_iterator<bool> (cout, " "));
        cout << endl ;
    }
private:
    // 按照活动结束时间非递减排序
    void __sortTime () 
    {
        sort (m_vTimeList.begin(), m_vTimeList.end()) ;
        for (vector<ActivityTime>::iterator ite = m_vTimeList.begin() ;
                ite != m_vTimeList.end() ; 
                ++ ite) {
            cout << ite->m_nStart << ", "<< ite ->m_nEnd << endl ;
        }
    }
private:
    vector<ActivityTime>    m_vTimeList ;    // 活动时间安排列表
    vector<bool>            m_bvSelectFlag ;// 是否安排活动标志
    int    m_nCount ;    // 总活动个数
} ;
int main()
{
    vector<ActivityTime> vActiTimeList ;
    vActiTimeList.push_back (ActivityTime(1, 4)) ;
    vActiTimeList.push_back (ActivityTime(3, 5)) ;
    vActiTimeList.push_back (ActivityTime(0, 6)) ;
    vActiTimeList.push_back (ActivityTime(5, 7)) ;
    vActiTimeList.push_back (ActivityTime(3, 8)) ;
    vActiTimeList.push_back (ActivityTime(5, 9)) ;
    vActiTimeList.push_back (ActivityTime(6, 10)) ;
    vActiTimeList.push_back (ActivityTime(8, 11)) ;
    vActiTimeList.push_back (ActivityTime(8, 12)) ;
    vActiTimeList.push_back (ActivityTime(2, 13)) ;
    vActiTimeList.push_back (ActivityTime(12, 14)) ;
    ActivityArrange aa (vActiTimeList) ;
    aa.greedySelector () ;
    return 0 ;
}

结果

带权活动选择问题

算法伪代码【核心算法】

问题描述：

会场出租：选择出租的活动时间不能冲突，怎样选择才能选更多的活动?

带权活动选择问题代码实现

package day1.java;
public class activityChoose {
    private static class Activity{
        int startTime;
        int endTime;
        int weight;
        private Activity(int startTime, int endTime, int weight){
            this.startTime = startTime;
            this.endTime = endTime;
            this.weight = weight;
        }
    }
    private static void activityChoose(Activity[] S){
        // 记录p：在a_i开始前最后结束的活动
        int[] p = new int[S.length+1];
        p[0] = 0;
        p[1] = 0;
        for(int i=2; i<=S.length; i++){
            for(int j=i-1; j>0; j--){
                if(S[j-1].endTime <= S[i-1].startTime){
                    p[i] = j;
                    break;
                }
            }
        }
        for(int i=1; i<=S.length; i++){
            System.out.println(p[i]);
        }
        int[] D = new int[S.length+1];
        int[] Rec = new int[S.length+1];
        D[0] = 0;
        // 动态规划
        for(int j=1; j<S.length+1; j++){
            if(D[p[j]]+S[j-1].weight > D[j-1]){
                D[j] = D[p[j]] + S[j-1].weight;
                Rec[j] = 1;
            }else{
                D[j] = D[j-1];
                Rec[j] = 0;
            }
        }
        // 输出方案
        int k=S.length;
        while(k > 0){
            if(Rec[k] == 1){
                System.out.println("选择:开始时间"+S[k-1].startTime+"结束时间"+S[k-1].endTime);
                k = p[k];
            }else{
                k--;
            }
        }
    }
    // 按结束时间从小到大排序
    private static void quickSortActivity(Activity[] S, int start, int end){
        int i = start;
        int j = end;
        if (start < end){
            Activity tmp = S[i];
            while(i<j){
                while(i<j && S[i].endTime <= S[j].endTime){
                    j--;
                }
                S[i] = S[j];
                while (i < j && S[i].endTime >= S[j].endTime) {
                    i++;
                }
                S[j] = S[i];
            }
            S[i] = tmp;
            quickSortActivity(S, start, i-1);
            quickSortActivity(S, i+1,end);
        }
    }
    public static void main(String[] args){
        Activity[] S = new Activity[10];
        S[0] = new Activity(1,4,1);
        S[1] = new Activity(3,5,6);
        S[2] = new Activity(0,6,4);
        S[3] = new Activity(4,7,7);
        S[4] = new Activity(3,9,3);
        S[5] = new Activity(5,9,12);
        S[6] = new Activity(6,10,2);
        S[7] = new Activity(8,11,9);
        S[8] = new Activity(8,12,11);
        S[9] = new Activity(2,14,8);
        quickSortActivity(S, 0, 9);
        activityChoose(S);
    }
}

结果