programming-challenges Shoemaker's Problem (110405) 题解

Greedy.

证明：

Let‘s say we have job 1, 2, ..., n, and they have time and fine as t1, f1, t2, f2, ..., tn, fn

and they are in the order of t1/f1 <= t2/f2 <= t3/f3 <= ... <= tn/fn

So this is the objective schedule. Now we change 1 with m (1 < m <= n)

By the original order, we need pay fine as much as: F1 = t1 * (f2 + ... + fn) + t2 * (f3 + ... + fn) + ... + tm * (fm+1 + ... + fn) + R

By the new order, we need pay fine as much as: F2 = tm * (f1 + ... + fm-1 + fm+1 + ... + fn) + t1 * (f2 + ... + fm-1 + fm+1 + ... + fn) + ... + fm-1 * fm+1 + ... + fn) + R

F1 - F2 = (t1 + t2 + ... + tm-1) * fm - (tm * f1 + tm * f2 + ... + tm * fm-1)

As t1 * fm <= tm * f1, t2 * fm <= tm * f2, ..., tm-1 * fm <= tm * fm-1 F1 - F2 <= 0

So the original order is the best order.

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <assert.h>
#include <algorithm>
#include <math.h>
#include <ctime>
#include <functional>
#include <string.h>
#include <stdio.h>
#include <numeric>
#include <float.h>

using namespace std;

/*
4.6.5
*/

struct Rec {
	int time, cost, id;
	Rec(int atime, int acost, int aid) : time(atime), cost(acost), id(aid) {}
};

bool camp(Rec r1, Rec r2) {
	return r1.time * r2.cost < r1.cost * r2.time;
}

int main() {
	int TC = 0; cin >> TC;
	bool blank = false;
	for (int tc = 1; tc <= TC; tc++) {
		int num = 0; cin >> num;
		vector<Rec> recs;
		for (int i = 0; i < num; i++) {
			int time, cost; cin >> time >> cost;
			recs.push_back(Rec(time, cost, i + 1));
		}

		stable_sort(recs.begin(), recs.end(), camp); 

		if (blank) {
			cout << endl;
		}
		blank = true;
		for (int i = 0; i < recs.size(); i++) {
			if (i > 0) cout << " ";
			cout << recs[i].id;
		}
		cout << endl;
	}
	return 0;
}

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programming-challenges Shoemaker's Problem (110405) 题解