C++与Java分别解决活动选择问题和带权活动选择问题

2024-04-02 19:04:59 222人浏览安东尼

摘要

目录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

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