UVA10026 Shoemaker's Problem【贪心】

Shoemaker has N jobs (orders from customers) which he must make. Shoemaker can work on only one job in each day. For each i-th job, it is known the integer Ti (1 ≤ Ti ≤ 1000), the time in days it takes the shoemaker to finish the job. For each day of delay before starting to work for the i-th job, shoemaker must pay a fine of Si (1 ≤ Si ≤ 10000) cents. Your task is to help the shoemaker, writing a programm to find the sequence of jobs with minimal total fine.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

First line of input contains an integer N (1 ≤ N ≤ 1000). The next N lines each contain two numbers: the time and fine of each task in order.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

You programm should print the sequence of jobs with minimal fine. Each job should be represented by its number in input. All integers should be placed on only one output line and separated by one space. If multiple solutions are possible, print the first lexicographically.

Sample Input

3 4

1 1000

2 2

5 5

Sample Output

2 1 3 4

题链接：UVA10026 Shoemaker's Problem

问题简述：

裁缝要修理N双鞋子，每天只能修一双，每个鞋子有规定修的天数以及每延迟一天修要罚钱的数目，找出花费罚钱最少的修鞋顺序。

问题分析：

这是一个经典贪心法解决的问题，令fine=money/day，每天罚钱fine越大，就要越早修理，使得总罚钱最少。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* UVA10026 Shoemaker's Problem */#include <bits/stdc++.h>using namespace std;const int N = 1000;
struct Job{int no;double day, money, fine;
} job[N];int cmp(Job a, Job b) {return fabs(a.fine - b.fine) > 0.00000001 ? a.fine > b.fine : a.no < b.no;
}int main()
{int t, n;scanf("%d", &t);while(t--) {scanf("%d", &n);for(int i = 0; i < n; i++) {scanf("%lf %lf", &job[i].day, &job[i].money);job[i].fine = job[i].money / job[i].day;job[i].no = i + 1;}sort(job, job + n, cmp);for (int i = 0; i < n - 1; i++)printf("%d ", job[i].no);printf("%d", job[n - 1].no);if (t)printf("\n\n");elseprintf("\n");}return 0;
}

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