Task
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16874 Accepted Submission(s): 4287

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 (500xi+2yi) 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
2014 Multi-University Training Contest 1

问题链接：HDU4864 Task
问题简述：（略）
问题分析：这是一个服务（机器提供服务）与任务的问题，用贪心法来解决，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* HDU4864 Task */

#include <bits/stdc++.h>

using namespace std;

const int N = 100000;
struct Node {
    int x, y;
} mach[N], task[N];
const int Y = 100 + 1;
int lvl[Y];

bool cmp(Node a, Node b)
{
    return a.x == b.x ? a.y > b.y : a.x > b.x;
}

int main()
{
    int n, m;
    while (~scanf("%d%d", &n, &m)) {
        for (int i = 0; i < n; i++)
            scanf("%d%d", &mach[i].x, &mach[i].y);
        for (int i = 0; i < m; i++)
            scanf("%d%d", &task[i].x, &task[i].y);

        sort(mach, mach + n, cmp);
        sort(task, task + m, cmp);

        long long sum = 0;
        int cnt = 0, j = 0;
        memset(lvl, 0, sizeof lvl);
        for (int i = 0; i < m; i++) {
            while (j < n && mach[j].x >= task[i].x) lvl[mach[j++].y]++;


            for (int k = task[i].y; k <= Y; k++)
                if (lvl[k]) {
                    cnt++;
                    sum += 500 * task[i].x + 2 * task[i].y;
                    lvl[k]--;
                    break;
                }
        }

        printf("%d %lld\n", cnt, sum);
    }

    return 0;
}