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; 
}