uva 10026 - Shoemaker's Problem

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本文介绍了一个经典的任务调度问题——鞋匠问题。鞋匠需要完成多项工作，每项工作有不同的完成时间和延误罚款。文章详细解释了如何通过比较工作单位时间罚款来确定最优的工作顺序，从而将总罚款降至最低，并提供了具体的实现代码。

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

Sample Output

2 1 3 4

方法：

本题题意为一个鞋匠有n件工作要做，每件工作都有其消耗时间和推辞一天开始做该工作所付出的代价。

求做这n件工作的顺序，使得总的代价最小。

假设只有两件工作，分别为J1和J2它们消耗的时间和推辞一天开始的代价分别为T1，F1和T2，F2.

若 F1/T1 < F2/T2，则T2 * F1 < T1 * F2 ，其中T2 * F1是J1先开始的代价，T1 * F2是J2先开始的代价。

所以要使代价最小，应该J2先开始。

反之，应使J1先开始。

若有n > 2，若Fi/Ti < Fi+1/Ti+1，则交换Ji和Ji+1的顺序就可使总代价变小则可采用插入排序的思想，所以以F/T大小按降序排序，就可得到所求的序列。

代码：

#include 
#define N 1000
typedef struct job job;
struct job{
    int no;
    double p;
};
void insertsort(job *job_arr_p[], int n)
{
    int i, j;
    job *t;
    for(i = 0; i < n; i++){
	t = job_arr_p[i];
	for(j = i-1; j >= 0 && job_arr_p[j]->p < t->p; j--)
	    job_arr_p[j+1] = job_arr_p[j];
	job_arr_p[j+1] = t;
    }
}
int main()
{
    int i, n, t;
    double time, fine;
    job job_arr[N], *job_arr_p[N];
    scanf("%d", &t);
    while(t-- >0){
	scanf("%d", &n);
	for(i = 0; i < n; i++)
	    job_arr_p[i] = job_arr + i;
	for(i = 0; i < n; i++){
	    scanf("%lf %lf", &time, &fine);
	    job_arr[i].p = fine/time;
	    job_arr[i].no = i + 1;
	}
	insertsort(job_arr_p, n);
	for(i = 0; i < n; i++)
	    printf("%d%s", job_arr_p[i]->no, i == n-1 ? "\n" : " ");
	if(t != 0)
	    printf("\n");
    }
    return 0;
}