HDU4864_Task_贪心

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Task

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6721 Accepted Submission(s): 1786

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

1 2100 3100 2100 1

Sample Output

1 50004

大致题意：

公司有不同类型的任务和不同类型的机器。任务和机器都有两个属性，时长和等级。只有机器的时长和等级都大于等于一项任务，机器才能完成这项任务。并且，一项任务只能由一台机器完成，一台机器只能完成一项任务。每完成一项任务的利润为 (500*xi+2*yi) 其中 xi yi 分别是任务的时间和等级。求完成任务的策略，使总利润最大化。

大体思路：

按时间由大到小，时间相同则按等级由高到低，将任务和机器分别排序。然后从第一个任务开始，挑选最合适的机器将其完成。最合适的原则是：时间大于该任务且等级最低。检索完所有的任务即可得到结果。

看到网上聚聚的思路，将时长符合要求的机器分成不同的等级保存在一个数组里。这样实际上简化了机器的数据，效率更高。

#include <stdio.h>  #include <string.h>  #include <algorithm>  using namespace std;    struct node  {      int x,y;  } s1[100005],s2[100005];    int cmp(node a,node b)  {      if(a.x == b.x)          return a.y>b.y;      return a.x>b.x;  }    int main()  {      int n,m,i,j,cnt;      __int64 sum;      while(~scanf("%d%d",&n,&m))      {          for(i = 0; i<n; i++)              scanf("%d%d",&s1[i].x,&s1[i].y);          for(i = 0; i<m; i++)              scanf("%d%d",&s2[i].x,&s2[i].y);          sort(s1,s1+n,cmp);          sort(s2,s2+m,cmp);          cnt = sum = 0;          int c[105] = {0};          for(i = 0,j = 0; i<m; i++)          {              while(j<n && s1[j].x>=s2[i].x)              {                  c[s1[j].y]++;                  j++;              }              for(int k = s2[i].y; k<=100; k++)              {                  if(c[k])                  {                      c[k]--;                      sum+=(s2[i].x*500+s2[i].y*2);                      cnt++;                      break;                  }              }          }          printf("%d %I64d\n",cnt,sum);      }  }