10026 - Shoemaker's Problem 贪心

weixin_30788619

于 2019-07-10 17:44:01 发布

阅读量209

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/DreamHighWithMe/p/3381606.html

博客围绕鞋匠的工作安排问题展开，鞋匠有N个工作，每天只能做一个，每个工作有完成时间Ti和延迟罚款Si。需编写程序找出使总罚款最小的工作顺序，还给出了输入输出的格式要求及示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Alex Gevak
September 16, 2000(Revised 4-10-00, Antonio Sanchez)

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<stack>
#include<set>
#include<iostream>
#include<vector>
#include<queue>
//ios_base::sync_with_stdio(false);
//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;
#define sz(v) ((int)(v).size())
#define rep(i, a, b) for (int i = (a); i < (b); ++i)
#define repf(i, a, b) for (int i = (a); i <= (b); ++i)
#define repd(i, a, b) for (int i = (a); i >= (b); --i)
#define clr(x) memset(x,0,sizeof(x))
#define clrs( x , y ) memset(x,y,sizeof(x))
#define out(x) printf(#x" %d\n", x)
#define sqr(x) ((x) * (x))
typedef long long LL;

const int INF = 1000000000;
const double eps = 1e-8;
const int maxn = 3000;

int sgn(const double &x) {  return (x > eps) - (x < -eps); }

struct Job
{
    int t;
    int f;
    int id;
}J[maxn];

bool cmp(Job j1,Job j2)
{
    return j1.t*j2.f < j1.f*j2.t;
}
int main() 
{
    //freopen("in.txt","r",stdin);

    int T;
    cin>>T;
    int first = 0;
    while(T--)
    {
        if(first)
            cout<<endl;
        first = 1;
        int n;
        cin>>n;
        repf(i,1,n)
        {
            scanf("%d%d",&J[i].t,&J[i].f);
            J[i].id = i;
        }
        
        sort(J+1,J+n+1,cmp);
        cout<<J[1].id;
        repf(i,2,n)
            cout<<" "<<J[i].id;
        cout<<endl;
        
    }
    return 0;
}