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 2 100 3 100 2 100 1

 

Sample Output

1 50004

 

题意：n台机器，m个任务，每台机器和任务都有两个值，机器的两个值都大于任务的两个值，这台机器才干完毕这个任务，每台机器仅仅能完毕一个任务，问最大收益

思路：贪心，标记好暴力就可以

#include 
      
       #include 
       
        #include 
        
         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
         
          =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);    }}