UVA 10026 Shoemaker's Problem

Shoemaker's Problem

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 T_i (1<=T_i<=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 S_i (1<=S_i<=10000) cents. Your task is to help the shoemaker, writing a programm to find the sequence of jobs with minimal
total fine.

The 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.

The 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

1

4
3 4
1 1000
2 2
5 5

Sample Output

2 1 3 4

题意：一个鞋匠接到非常多订单。可是，每一个客户都觉得自己的订单应该被立即处理。因此，对于第i个订单。在開始处理这个订单之前，每天都要付罚金Si (1<=Si≤<=1000)。而他一天仅仅能处理一个订单。并且一个订单可能须要非常多天才干完毕。对于第i个订单。整数Ti (1<=Ti<=1000)代表处理完毕这个订单所须要的天数。

求所付罚金最少的订单处理顺序。

分析：贪心解决。对每一个订单 罚金/天数 从大到小排序。结果一样按序号排序（保证字典序最小）。

证明：假设x和y为排好的顺序中相邻的两个订单，因为x、y之后的订单顺序是固定的。所以不管是排成xy还是排成yx，对后面的罚金没有影响。罚金区别在于是排成xy还是yx。

假设是xy,则罚金为Tx*Sy；假设是yx，则罚金是Ty*Sx。

假设Tx*Sy<Ty*Sx，就排成xy；否则排成yx。

所以这样的贪心策略是正确的。

#include<cstdio>
#include<algorithm>
using namespace std;

struct shoe {
    int id;
    int time;
    int fine;
} a[1005];

bool comp(shoe x, shoe y) {  //用乘法比較。避免相除以后浮点数产生误差
    if(x.fine * y.time != x.time * y.fine)
        return x.fine * y.time > x.time * y.fine;
    return x.id < y.id;
}

int main()
{
    int T, n, i;
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        for(i = 0; i < n; i++) {
            scanf("%d%d",&a[i].time, &a[i].fine);
            a[i].id = i + 1;
        }
        sort(a, a+n, comp);
        for(i = 0; i < n - 1; i++)
            printf("%d ", a[i].id);
        printf("%d\n", a[n-1].id);
        if(T > 0) printf("\n");
    }
    return 0;
}