hdu4864Task(馋)

主题链接：

啊哈哈。点我

题目：

Task

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2512    Accepted Submission(s): 643

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

2014 Multi-University Training Contest 1

 

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思路：

首先考虑获得的酬劳。。５００＊ｘｉ＋２＊ｙｉ，所以yi能够当做次要因素，主观因素是时间。所以对任务。和机器的时间大- >小，等级大->小排序。。

接下来就是枚举，将全部满足机器执行时间》=任务时间的增加数组。然后选满足完毕任务的最小等级的机器去完毕任务。

这种巧妙之处在于后面的加进来的任务必然能够被先前加进来的机所完毕。由于任务是按时间降序排列的。。

那样这道题就得到了完美的解决。。

代码为：

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

const int maxn=100000+10;
int level[100+10];
int n,m,sum;
__int64 ans;

struct P
{
    int xi,yi;
}machine[maxn],task[maxn];

bool cmp(P a,P b)
{
   if(a.xi==b.xi) return a.yi>b.yi;
   return a.xi>b.xi;
}

void read_data()
{
    for(int i=1;i<=n;i++)
        scanf("%d%d",&machine[i].xi,&machine[i].yi);
    for(int i=1;i<=m;i++)
        scanf("%d%d",&task[i].xi,&task[i].yi);
    sort(task+1,task+1+m,cmp);
    sort(machine+1,machine+1+n,cmp);
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        ans=sum=0;
        read_data();
        memset(level,0,sizeof(level));
        for(int i=1,j=1;i<=m;i++)
        {
            while(j<=n&&machine[j].xi>=task[i].xi)
            {
                level[machine[j].yi]++;
                j++;
            }
            for(int k=task[i].yi;k<=100;k++)
            {
                if(level[k])
                {
                    level[k]--;
                    ans=ans+500*task[i].xi+2*task[i].yi;
                    sum++;
                    break;
                }
            }
        }
        printf("%d %I64d\n",sum,ans);
    }
    return 0;
}

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