洛谷P2240 Java解法

题目出处点这里
思路：构造一个金币类，按性价比从小到大进行排序即可，不过要注意可能有全部金币都取完了但是背包还没满的情况，此时直接输出maxPrice即可

代码如下（很好理解）：

package greedy;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;

public class P2240 {

	static ArrayList<coin> list = new ArrayList<coin>();
	static double maxPrice = 0;
	static double curWeight = 0;
	static int index;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		double N = sc.nextInt();
		double T = sc.nextInt();
		for (int i = 0; i < N; i++) {
			list.add(new coin(sc.nextInt(), sc.nextInt()));
		}
		//这里需要自己写一个比较器，只对coin对象的xjb进行由大到小的排序
		Collections.sort(list, new Comparator<coin>() {
			public int compare(coin a, coin b) {
				double s = (b.xjb - a.xjb);
				if (s > 0) {
					return 1;
				} else if (s == 0) {
					return 0;
				} else {
					return -1;
				}
			}
		});
		for (int i = 0; i < N; i++) {
			if (curWeight + list.get(i).w > T) {
				break;
			}
			curWeight = curWeight + list.get(i).w;
			maxPrice = maxPrice + list.get(i).p;
			index = i;
		}

		double restWeight = T - curWeight;// 剩余的容量

		if (restWeight != 0 && index + 1 < N) {
			maxPrice = maxPrice + restWeight * list.get(index + 1).p / list.get(index + 1).w;
			System.out.printf("%.2f", maxPrice);
		} else {// 注意当index + 1 == N 时,就说明东西都放进背包里了,此时有可能背包没放满,也有可能刚刚放满
			System.out.printf("%.2f", maxPrice);
		}

	}
}

class coin {

	double w;// 重量
	double p;// 价值
	double xjb;// 性价比

	// 构造器
	public coin(double w, double p) {
		this.w = w;
		this.p = p;
		this.xjb = p / w;
	}
}