Task

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2154    Accepted Submission(s): 549


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 題意：

題目中告訴我們有n臺機器和m個任務，每一臺機器一天可以解決一個任務，一個任務一天也僅能由一臺機器完成。對於它們之間的關係必須滿足機器的執行時間比任務的工作時間長，以及機器的工作等級比任務的工作等級高。
當完成一個任務時，就可以得到 “500 * 任務時間 + 2 * 任務等級” 的金錢，問最多能完成多少任務同時得到最多的金錢。

思路：

首先這裡的機器和任務是一對一的匹配，為了取得最多的金錢考慮到先去解決可以盈利較多的任務，那麼這裡就對任務和機器進行排序，採取先大後小的策略，由於任務時間佔的權重較大(500 : 2)，所以優先排序時間後排序等級。
對於每一個任務，每次取到最優的機器去做該任務。若採用全部重新遍歷機器的方式，必定會超時(N * M)。
考慮到能完成前面任務（時間長，等級高的任務）的機器對於時間方面一定能達到要求。所以將這些能完成前面任務的機器全部標記下來，然後從中找出等級最小且滿足等級要求的機器去完成任務。

/*************************************************************************
	> File Name: hdu4864.cpp
	> Author: Bslin
	> Mail: [email protected]
	> Created Time: 2014年07月23日 星期三 18時40分14秒
 ************************************************************************/

#include <cstdio>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 100010

struct node {
	int time, level;

	bool operator < (const node &b) const {
		if(time == b.time) {
			return level > b.level;
		}
		return time > b.time;
	}
} mach[N], task[N];
int vis[110];

int main(int argc, char *argv[]) {
	freopen("in.txt", "r", stdin);
	int n, m, i, j, k;
	long long ans, num;
	while(scanf("%d%d", &n, &m) != EOF) {
		for (i = 0; i < n; ++i) {
			scanf("%d%d", &mach[i].time, &mach[i].level);
		}
		for (i = 0; i < m; ++i) {
			scanf("%d%d", &task[i].time, &task[i].level);
		}
		sort(mach, mach + n);
		sort(task, task + m);
		memset(vis, 0, sizeof(vis));
		ans = 0;
		num = 0;
		for (i = 0, j = 0; i < m; ++i) {
			while(j < n && mach[j].time >= task[i].time) {
				vis[mach[j].level] ++;
				j ++;
			}
			for (k = task[i].level; k <= 100; ++k) {
				if(vis[k]) {
					num ++;
					vis[k] --;
					ans += 500 * task[i].time + 2 * task[i].level;
					break;
				}
			}
		}
		printf("%I64d %I64d\n", num, ans);
	}
	return 0;
}