HDU 4864----Task

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.

 

典型贪心问题，求满足条件的最优解

例子    300 7   270 5

        270 5   260 5

250 5

这个题的核心是如果从0开始暴力枚举的话会超时且找不到最优解，用到贪心的思想，枚举任务，把每次满足任务的机器存到c数组里，比如例子，指针j走到270的地方break，c数组里面存的c[5]=1,c[7]=1，然后开始从任务需要的等级开始找发现c[5]可以，然后计算270 5 这时候300 7(7更大)可以留下为了装下面更大的数，c[5]--

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include<cmath>
#define INF 0x7f7f7f7f
#include<vector>
#include<string>
#include<queue>
#include<map>
#define N  605
#define inf 100
using namespace std;
struct point
{
    int x,y;
} a[100005],b[100005];
int vis[101];
bool cmp(struct point a,struct point b)
{
    if(a.x==b.x)
        return a.y>b.y;
    return a.x>b.x;
}
int main()
{
    int i,j,k,l,m,n,p;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0; i<n; i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        for(i=0; i<m; i++)
            scanf("%d%d",&b[i].x,&b[i].y);
        memset(vis,0,sizeof(vis));
        sort(a,a+n,cmp);
        sort(b,b+m,cmp);
        __int64 s=0;
        __int64 t=0;
        for(i=0,j=0; i<m; i++)
        {
            for(j; j<n; j++)//如果从0开始找满足条件会超时
            {
                if(a[j].x>=b[i].x)
                    vis[a[j].y]++;
                else
                    break;
            }
            for(k=b[i].y; k<101; k++)
            {
                if(vis[k])
                {
                    t++;
                    s+=(b[i].x*500+2*b[i].y);
                    vis[k]--;
                    break;
                }
            }
        }
        printf("%I64d %I64d\n",t,s);
    }
}