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(Java实现) 活动选择

作者:互联网

活动选择的类似问题都可以这么写

import java.util.ArrayList;
public class huodongxuanze {

	    /**
	     * //算法导论中活动选择问题动态规划求解
	     * @param s 活动的开始时间
	     * @param f 活动的结束时间
	     * @param n 活动数目
	     * @return 最大兼容的活动个数
	     */
	    public static int maxCompatiableActivity(int[] s, int[] f, int n){
	        int[][] c = new int[n + 2][n + 2];
	        
	        for(int j = 0; j <= n+1; j++)
	            for(int i = n+1; i >= j; i--)
	                c[i][j] = 0;//if i>=j S(i,j)是空集合
	        
	        int maxTemp = 0;
	        for(int j = 1; j <= n+1; j++)
	        {
	            for(int i = 0; i < j; i++)//i < j
	            {
	                for(int k = i+1; k < j; k++)// i< k <j
	                {
	                    if(s[k] >= f[i] && f[k] <= s[j])//S(i,j)不空
	                    {
	                        if(c[i][k] + c[k][j] + 1 > maxTemp)
	                            maxTemp = c[i][k] + c[k][j] + 1;
	                    }
	                }//inner for
	                c[i][j] = maxTemp;
	                maxTemp = 0;
	            }//media for
	        }//outer for
	        return c[0][n+1];
	    }
	    
	    //贪心算法的递归解
	    public static ArrayList<Integer> greedyActivitySelection(int[] s, int[] f, int i, int n, ArrayList<Integer> activities){
	        //初始调用时 i = 0, 所以a(1)是必选的(注意:活动编号已经按结束时间排序)
	        int m = i + 1;
	        
	        //s[m] < f[i] 意味着活动 a(m) 与 a(i)冲突了
	        while(m <= n && s[m] < f[i])
	            m++;//选择下一个活动
	        
	        if(m <= n){
	            activities.add(m);
	            greedyActivitySelection(s, f, m, n, activities);
	        }
	        return activities;
	    }
	    
	    //贪心算法的非递归解, assume f[] has been sorted and actId 0/n+1 is virtually added
	    public static ArrayList<Integer> greedyActivitySelection2(int[] s, int[] f, int n, ArrayList<Integer> acitivities){
	        //所有真正的活动(不包括 活动0和 活动n+1)中,结束时间最早的那个活动一定是最大兼容活动集合中的 活动.
	        int m = 1;
	        acitivities.add(m);
	        
	        for(int actId = 2; actId <= n; actId++){
	            if(s[actId] >= f[m])//actId的开始时间在 m 号活动之后.--actId 与 m 没有冲突
	            {
	                m = actId;
	                acitivities.add(m);
	            }
	        }
	        return acitivities;
	    }
	    
	    //for test purpose
	    public static void main(String[] args) {
	        //添加了 a(0) 和 a(n+1)活动. 其中s(0)=f(0)=0, s(n+1)=f(n+1)=Integer.MAX_VALUE
	        int[] s = {0,1,3,0,5,3,5,6,8,8,2,12,Integer.MAX_VALUE};//start time
	        int[] f = {0,4,5,6,7,8,9,10,11,12,13,14,Integer.MAX_VALUE};//finish time
	        int n = 11;//活动的个数
	        int result = maxCompatiableActivity(s, f, n);
	        System.out.println("最大兼容活动个数: " + result);
	        
	        ArrayList<Integer> acts = new ArrayList<Integer>();
	        greedyActivitySelection(s, f, 0, n, acts);
	        for (Integer activityId : acts)
	            System.out.print(activityId + " ");
	        
	        System.out.println();
	        ArrayList<Integer> acts2 = new ArrayList<Integer>();
	        greedyActivitySelection2(s, f, n, acts2);
	        for (Integer activityId : acts2)
	            System.out.print(activityId + " ");
	    }
	}


标签:maxTemp,Java,int,ArrayList,选择,Integer,活动,actId
来源: https://blog.csdn.net/a1439775520/article/details/90729227