hdu 4864 Task【贪心】

Task

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5241    Accepted Submission(s): 1365

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

 

题目大意：有n个机器有m个任务，需要机器x和y都要大于等于任务的x，y就能得到500*x+2*y的任务报酬，问最多能够解决多少任务，一共能够获得多少报酬。

解题思路：排序任务的x，排序机器的x，然后for一遍任务，在机器中找符合条件的xj>=xi。然后在所有符合条件的xj中，找到yj最小的，作为解决这个任务的机器。

AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
struct task
{
    int x,y;
    friend bool operator <(task a,task b)
    {
        return a.y<b.y;
    }
}a[100005],b[100005];
int vis[120];
int cmp(task a,task b)
{
    if(a.x!=b.x)return a.x>b.x;
    else return a.y>b.y;
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(vis,0,sizeof(vis));
        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);
        }
        sort(a,a+n,cmp);
        sort(b,b+m,cmp);
        memset(vis,0,sizeof(vis));
        int j=0;
        int cont=0;
        __int64 output=0;
        for(int i=0;i<m;i++)
        {
            while(j<n&&a[j].x>=b[i].x)
            {
                vis[a[j].y]++;
                j++;
            }
            for(int k=b[i].y;k<=100;k++)
            {
                if(vis[k]>0)
                {
                    cont++;
                    vis[k]--;
                    output+=500*b[i].x+2*b[i].y;
                    break;
                }
            }
        }
        printf("%d %I64d\n",cont,output);
    }
}