8.5贪心策略例题：区间选点问题

// 数轴上有n个闭区间[ai,bi]。取尽量少的点，使得每个区间内都至少有一个点（不同区间内含的点可以是同一个）。
/*
Intervals
You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn.
Write a program that:
reads the number of intervals, their end points and integers c1, ..., cn from the standard input,
computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,...,n,
writes the answer to the standard output.

Input
The first line of the input contains an integer n (1 <= n <= 50000) -- the number of intervals.
The following n lines describe the intervals. The (i+1)-th line of the input contains three integers ai,
bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.

Output
The output contains exactly one integer equal to the minimal size of set Z
sharing at least ci elements with interval [ai, bi], for each i=1,2,...,n.
Sample Input
5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1
Sample Output
6
 */
//POJ1201

思路：
每次先选单个区间最后一个点，这样更容易包含其它区间，取尽量少的点

import java.util.Arrays;
import java.util.Scanner;

public class Eight_5贪心策略例题_区间选点问题 {
    private static class Interval implements Comparable<Interval>{
        int s;
        int t;
        int c;
        
        public Interval(int s, int t, int c) {
            this.s = s;
            this.t = t;
            this.c = c;
        }

        @Override
        public int compareTo(Interval other) {
            int x = this.t - other.t;
            if(x == 0)
                return this.s - other.s;
            else
                return x;
        }
    }
    
    /**
     * 统计axis上s-t区间已经有多少个点被选中
     */
    private static int sum(int[] axis, int s, int t){
        int sum = 0;
        for(int i = s; i <= t; i++){
            sum += axis[i];
        }
        return sum;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        Interval[] intervals = new Interval[n];
        for(int i = 0; i < n; i++){
            intervals[i] = new Interval(in.nextInt(), in.nextInt(), in.nextInt());
        }
        Arrays.sort(intervals);    //按区间右端点排序
        
        int max = intervals[n-1].t;    //右端最大值
        int[] axis = new int[max+1];    //标记数轴上的点是否已经被选中
        //int[] sums = new int[max+1];
        for(int i = 0; i < n; i++){
            //1.查阅区间中有多少个点
            int s = intervals[i].s; //起点
            int t = intervals[i].t; //终点
            int cnt = sum(axis, s, t);    //找到这个区间已经选点的数量，sum[t]-sum[s-1];低效率
                
            //2.如果不够，从区间右端开始标记，遇标记过的就跳过
            intervals[i].c -= cnt; //需要新增的点的数量
            while(intervals[i].c > 0){
                if(axis[t] == 0){ //从区间终点开始选点
                    axis[t] = 1;
                    intervals[i].c--;    //进一步减少需要新增的数量
                    t--;
                }else{    //如果这个点已经被选过了
                    t--;
                }
            }
        }
        System.out.println(sum(axis, 0, max));
    }
}