Task

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
Author FZU
Source
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4906 4904 4903 4902 4901  題目大意：

有n臺機器，m個任務，每臺機器有xi,yi，每個任務也有xj,yj，當一個任務可以被處理的條件是，xj<=xi 且 yj<yi，處理完產生 500*xj+2*yj 的價值，問你最多產生的價值是多少？

解題思路：

注意y的範圍是 y<100，也就是x相差1,y不管相差多少價值都很少。

根據貪心的做法，肯定從高價值物品生產也就是按x排好序，再貪心，高價值的物品只需要在x比它大的所有機器中選擇y滿足條件的最小的那個（這個思考一下）

解題程式碼：

#include <iostream>
#include <set>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn=110000;

struct node{
    int x,y;
    friend bool operator < (node a, node b){
        if(a.y!=b.y) return a.y<b.y;
        else return a.x<b.x;
    }
};

int n,m;
node a[maxn],b[maxn];

bool cmp(node a,node b){
    if(a.x!=b.x) return a.x>b.x;
    else return a.y>b.y;
}

void solve(){
    multiset <node> mys;
    sort(a,a+n,cmp);
    sort(b,b+m,cmp);
    int r=0,cnt=0;
    long long ans=0;
    for(int i=0;i<m;i++){
        while(r<n && a[r].x>=b[i].x) mys.insert(a[r++]);
        multiset <node>::iterator it=mys.lower_bound(b[i]);
        if(it!=mys.end()){
            mys.erase(it);
            cnt++;
            ans+=b[i].x*500+b[i].y*2;
        }
    }
    cout<<cnt<<" "<<ans<<endl;
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=0;i<n;i++) scanf("%d%d",&a[i].x,&a[i].y);
        for(int i=0;i<m;i++) scanf("%d%d",&b[i].x,&b[i].y);
        solve();
    }
    return 0;
}