HDU 4864 Task

Task

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7281    Accepted Submission(s): 1922

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

解题思路：纯粹的贪心，在所有的机器中的找出刚好能完成某个任务的机器（从时间和level两个方面考虑），把所有的任务扫一遍。

代码如下:

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct Node{int xi,yi;}mm[100005],tt[100005];int cmp(Node a,Node b){if(a.xi==b.xi)return a.yi>b.yi;return a.xi>b.xi;}int main(){int i,j;int n,m;long long sum,cnt;while(~scanf("%d%d",&n,&m)){sum=cnt=0;for(i=0;i<n;i++)scanf("%d%d",&mm[i].xi,&mm[i].yi);for(i=0;i<m;i++)scanf("%d%d",&tt[i].xi,&tt[i].yi);sort(mm,mm+n,cmp);sort(tt,tt+m,cmp);int vi[105]={0};for(i=0,j=0;i<m;i++){while(j<n && mm[j].xi>=tt[i].xi){vi[mm[j].yi]++;j++;}for(int k=tt[i].yi;k<=100;k++){if(vi[k]){vi[k]--;                    sum+=(tt[i].xi*500+tt[i].yi*2);                      cnt++;                      break;}}}printf("%lld %lld\n",cnt,sum);}return 0;}