hdu4864 Task 2014 Multi-University Training Contest 1

Task

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 259    Accepted Submission(s): 47

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

 

Source

2014 Multi-University Training Contest 1

 

Recommend

We have carefully selected several similar problems for you:  4871 4870 4869 4868 4867 

贪心的思想，任务和机器都按时间由大到小 时间相同等级由大到小排列，然后从任务开始   每次找出等级大于他的机器存进数组，在数组里取等级最小的用即可。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define inf 0x3f3f3f3ftypedef struct node{    int u,v;} nodepoi;nodepoi t1[111111],t2[111111];int c[111];int n,m,u,v;int cmp(node a,node b){    if(a.u==b.u)return a.v>b.v;    return a.u>b.u;}int main(){    while(scanf("%d%d",&n,&m)!=EOF){        long long  sum=0,num=0;        memset(t1,0,sizeof(t1));        memset(t2,0,sizeof(t2));        memset(c,0,sizeof(c));        for(int i=0; i<n; i++){            scanf("%d%d",&u,&v);            t1[i].u=u;            t1[i].v=v;        }        for(int i=0; i<m; i++){            scanf("%d%d",&u,&v);            t2[i].u=u;            t2[i].v=v;        }        sort(t1,t1+n,cmp);        sort(t2,t2+m,cmp);        for(int i=0,j=0; i<m; i++){            while(j<n&&t1[j].u>=t2[i].u){                c[t1[j].v]++;                j++;            }            for(int k=t2[i].v; k<=100; k++){                if(c[k]){                    num++;                    c[k]--;                    sum+=500*t2[i].u+2*t2[i].v;                    break;                }            }        }        cout<<num<<' '<<sum<<endl;    }    return 0;}