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2013
11-09

POJ 1337 A Lazy Worker [解题报告] Java

A Lazy Worker

问题描述 :

There is a worker who may lack the motivation to perform at his peak level of efficiency because he is lazy. He wants to minimize the amount of work he does (he is Lazy, but he is subject to a constraint that he must be busy when there is work that he can do.)

We consider a set of jobs 1, 2,…, n having processing times t1, t2,…,tn respectively. Job i arrives at time ai and has its deadline at time di. We assume that ti, ai, and di have nonnegative integral values. The jobs have hard deadlines, meaning that each job i can only be executed during its allowed interval Ii=[ai, di]. The jobs are executed by the worker, and the worker executes only one job at a time. Once a job is begun, it must be completed without interruptions. When a job is completed, another job must begin immediately, if one exists to be executed. Otherwise, the worker is idle and begins executing a job as soon as one arrives. You should note that for each job i, the length of Ii, di – ai, is greater than or equal to ti, but less than 2*ti.

Write a program that finds the minimized total amount of time executed by the worker.

输入:

The input consists of T test cases. The number of test cases (T ) is given in the first line of the input file. The number of jobs (0<=n<=100) is given in the first line of each test case, and the following n lines have each job's processing time(1<=ti<=20),arrival time(0<=ai<=250), and deadline time (1<=di<=250) as three integers.

输出:

Print exactly one line for each test case. The output should contain the total amount of time spent working by the worker.

样例输入:

3
3
15 0 25
50 0 90
45 15 70
3
15 5 20
15 25 40
15 45 60
5
3 3 6
3 6 10
3 14 19
6 7 16
4 4 11

样例输出:

50
45
15

解题代码:

//* @author: 
import java.util.*;
public class Main {
  

 public static void main(String[] args){

    Scanner in = new Scanner(System.in);
	
   int ans[]=new int[300];
   int a[]=new int[100],d[]=new int[100],t[]=new int[100];

   int cas, i, j, n, temp;
   boolean key;
   cas=in.nextInt();

   while(( cas-- )!=0)
   {
    n=in.nextInt();
    for( i=0; i< n; i++ ){
      t[i]=in.nextInt();
      a[i]=in.nextInt();
      d[i]=in.nextInt();
    }
		
   for( i=0; i<=250; i++ )
 	ans[i] = 0;

  for( i=249; i>=0; i-- )
  {
	key = true;ans[i] = 99999999;
	for( j=0; j< n; j++ )
	{
         if( a[j] <= i && i+t[j] <= d[j] && ( temp = ans[i+t[j]] + t[j] ) < ans[i] ){
		ans[i] = temp;
              key = false;
         }
	}
	if( key ) ans[i] = ans[i+1];
		
   }
   System.out.printf( "%d\n", ans[0] );
  }
 }
}

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  3. #include <stdio.h>
    int main(void)
    {
    int arr[] = {10,20,30,40,50,60};
    int *p=arr;
    printf("%d,%d,",*p++,*++p);
    printf("%d,%d,%d",*p,*p++,*++p);
    return 0;
    }

    为什么是 20,20,50,40,50. 我觉得的应该是 20,20,40,40,50 . 谁能解释下?

  4. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮

  5. #include <cstdio>
    #include <cstring>

    const int MAXSIZE=256;
    //char store[MAXSIZE];
    char str1[MAXSIZE];
    /*
    void init(char *store) {
    int i;
    store['A']=’V', store['B']=’W',store['C']=’X',store['D']=’Y',store['E']=’Z';
    for(i=’F';i<=’Z';++i) store =i-5;
    }
    */
    int main() {
    //freopen("input.txt","r",stdin);
    //init(store);
    char *p;
    while(fgets(str1,MAXSIZE,stdin) && strcmp(str1,"STARTn")==0) {
    if(p=fgets(str1,MAXSIZE,stdin)) {
    for(;*p;++p) {
    //*p=store[*p]
    if(*p<’A’ || *p>’Z') continue;
    if(*p>’E') *p=*p-5;
    else *p=*p+21;
    }
    printf("%s",str1);
    }
    fgets(str1,MAXSIZE,stdin);
    }
    return 0;
    }

  6. 题目需要求解的是最小值,而且没有考虑可能存在环,比如
    0 0 0 0 0
    1 1 1 1 0
    1 0 0 0 0
    1 0 1 0 1
    1 0 0 0 0
    会陷入死循环