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本文介绍了一种用于优化公司任务分配的算法。该算法通过贪心策略实现，在满足机器工作时间和等级限制的前提下，最大化可完成的任务数量及收益。文章提供了具体实现代码。

Task

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1369 Accepted Submission(s): 341

Problem Description

Today the company has m tasks to complete. The ith task need xi minutes to complete. Meanwhile, this task has a difficulty level yi. The machine whose level below this task’s level yi cannot complete this task. If the company completes this task, they will get (500*xi+2*yi) dollars.
The company has n machines. Each machine has a maximum working time and a level. If the time for the task is more than the maximum working time of the machine, the machine can not complete this task. Each machine can only complete a task one day. Each task can only be completed by one machine.
The company hopes to maximize the number of the tasks which they can complete today. If there are multiple solutions, they hopes to make the money maximum.

Input

The input contains several test cases.
The first line contains two integers N and M. N is the number of the machines.M is the number of tasks(1 < =N <= 100000,1<=M<=100000).
The following N lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the maximum time the machine can work.yi is the level of the machine.
The following M lines each contains two integers xi(0<xi<1440),yi(0=<yi<=100).xi is the time we need to complete the task.yi is the level of the task.

Output

For each test case, output two integers, the maximum number of the tasks which the company can complete today and the money they will get.

Sample Input

Sample Output

  
   1 50004

Author

FZU

Source

2014 Multi-University Training Contest 1

题目链接点击打开链接

本题的策略是贪心，

对于价值c=500*xi+2*yi，yi最大影响100*2<500,所以就是求xi总和最大。可以先对机器和任务的时间从大到小排序。从最大时间的任务开始，找出满足任务时间要求的所有机器，从中找出等级最低且满足任务等级要求的机器匹配。依次对任务寻找满足要求的机器

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct s{
int x,y;
}a[100010],b[100010];
int cmp(const void *a,const void *b)
{
if((*(struct s *)a).x==(*(struct s *)b).x)
  return (*(struct s *)b).y-(*(struct s *)a).y;
else
  return (*(struct s *)b).x-(*(struct s *)a).x;
}
int main()
{
int m,n,d[110];
while(scanf("%d%d",&n,&m)!=EOF)
{
  int i,j=0,k;
  __int64 c=0,sum=0;
  memset(d,0,sizeof(d));
  //memset(v,0,sizeof(v));
  for(i=0;i<n;i++)
  {
   scanf("%d%d",&a[i].x,&a[i].y);
   //a[i].v=0;
  }
  for(i=0;i<m;i++)
  {
   scanf("%d%d",&b[i].x,&b[i].y);
   //b[i].v=0;
  }
  qsort(a,n,sizeof(a[0]),cmp);
  qsort(b,m,sizeof(b[0]),cmp);
  /*for(i=0;i<m;i++)
   printf("%d %d ",b[i].x,b[i].y);*/
  for(i=0;i<m;i++)
  {
   while(j<n&&a[j].x>=b[i].x)
   {
    d[a[j].y]++;//统计等级数
    j++;//j无需从0开始遍历，因为已经统计好等级数，且是按时间排的，比上一个任务时间大，必定比现在一个任务时间大
   }
   for(k=b[i].y;k<=100;k++)
   {
    if(d[k])
    {
     sum+=500*b[i].x+2*b[i].y;
     c++;
     d[k]--;
     //v[j]=0;
     break;
    }
   }
  }
  printf("%I64d %I64d\n",c,sum);
}
}