hdu 4864 排序+枚举

Task

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4580    Accepted Submission(s): 1202

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 2100 3100 2100 1

 

Sample Output

1 50004

分析：定义一个结构体，用来存放机器或任务的时间的等级，然后分别建立两个结构体数组，分别存放机器和任务。排序优先按照时间排序，在按照等级排序，工作时间长，等级高的机器当然是来完成时间长等级高的任务，接下来依次枚举每个任务，把可以完成当前任务时间的机器保存起来，然后从当前任务的等级开始枚举，看一下保存在数组中的是否存在一个机器的等级>=任务等级，如果存在，那么就计算这个任务。保存机器的数组要减去这台机器，因为已经用过了。然后再去枚举下一个任务

#include<iostream>#include<string>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>#define LL long longusing namespace std;const int maxn=100005;int N,M;struct node{    int x;    int y;};bool cmp(node a,node b){    if(a.x==b.x)return a.y>b.y;    return a.x>b.x;}int main(){    while(scanf("%d%d",&N,&M)!=EOF){        node mac[maxn],tak[maxn];        for(int i=0;i<N;i++)            scanf("%d%d",&mac[i].x,&mac[i].y);        for(int i=0;i<M;i++)            scanf("%d%d",&tak[i].x,&tak[i].y);        sort(mac,mac+N,cmp);        sort(tak,tak+M,cmp);        int a[105];        LL sum=0,cnt=0;        memset(a,0,sizeof(a));        for(int i=0,j=0;i<M;i++){            while(j<N&&mac[j].x>=tak[i].x){                a[mac[j].y]++;                j++;            }            for(int k=tak[i].y;k<101;k++){                if(a[k]){                    a[k]--;                    cnt++;                    sum+=500*tak[i].x+2*tak[i].y;                    break;                }            }        }        cout<<cnt<<' '<<sum<<endl;    }}