HDU 4864 Task(贪心)

Task

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5255 Accepted Submission(s): 1370

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

Author

FZU

Source

2014 Multi-University Training Contest 1 

/*思路：获得的价钱为500*xi+2*yi 所以获利最大就是500*xi+2*yi越大的任务由于题目范围(0<xi<1440),yi(0=<yi<=100)假设xi=2,yi=1与xj=1,yj=100  时间只要多一，1*500也会大于100*2； 所以排序方法为时间优先，再排序level 对任务和机器进行排序  枚举每一个任务，扫描机器从大到小，可以就完成他，累计获利，并标记这个机器被搞了 *//*超时！！  枚举机器的姿势不对！要用level数组标记，复杂度大大减少O(100)如果枚举机器 O(n) #include <iostream>#include <string.h>#include <algorithm>#include <cstdio>#include <math.h>#include <cstring>using namespace std;const int N=100000+10;struct Node{int time,level;}machine[N],task[N];bool cmp(Node a,Node b){    //排序    if(a.time==b.time)       return a.level>b.level;return a.time>b.time;} bool book[N];      //标记机器 int main(){     int n,m,i,j,sum;    long long re;  //要用long long     while(scanf("%d%d",&n,&m)==2){    re=0;    sum=0;    for(i=0;i<n;i++)    {  book[i]=0;  scanf("%d%d",&machine[i].time,&machine[i].level);  //cin>>machine[i].time>>machine[i].level;}                for(i=0;i<m;i++)     scanf("%d%d",&task[i].time,&task[i].level);     //  cin>>task[i].time>>task[i].level;sort(machine,machine+n,cmp);sort(task,task+m,cmp);for(i=0;i<m;i++){  //枚举每一个任务       int id=-1;   //标记机器位置    int tmp=99999999;   //标记机器level 初始为最大 选择level小的机器 for(j=0;j<n;j++){      //枚举每一个机器 若可以 还要选择等级最小的机器去完成   if(machine[j].time>=task[i].time) //这里要放在外面   {  //注意 wa了很久   //此条件要写在里面，因为如果机器被使用或由于等级不够，//  就被跳出了，后面没考虑到   if(machine[j].level>=task[i].level&&book[j]!=1  &&machine[j].level<=tmp){ id=j;      tmp=machine[j].level;  }        }  else   //跳出的条件是机器时间不足以完成这个任务 后面才不考虑   break;    }if(id!=-1){  //找到 book[id]=1;sum++;re+=task[i].time*500+task[i].level*2;}}//cout<<sum<<" "<<re<<endl; printf("%d %d\n",sum,re);}return 0;}*/#include <iostream>#include <string.h>#include <algorithm>#include <cstdio>#include <math.h>#include <cstring>using namespace std;const int N=100000+10;struct Node{int time,level;}machine[N],task[N];bool cmp(Node a,Node b){    //排序    if(a.time==b.time)       return a.level>b.level;return a.time>b.time;} bool book[N];      //标记机器 int main(){     int n,m,i,j,sum;    long long re;  //要用long long     while(scanf("%d%d",&n,&m)==2){    re=0;    sum=0;    for(i=0;i<n;i++)    {  book[i]=0;  scanf("%d%d",&machine[i].time,&machine[i].level);  //cin>>machine[i].time>>machine[i].level;}                for(i=0;i<m;i++)     scanf("%d%d",&task[i].time,&task[i].level);     //  cin>>task[i].time>>task[i].level;sort(machine,machine+n,cmp);sort(task,task+m,cmp);int level[105]={0};for(i=0,j=0;i<m;i++){  //枚举每一个任务  //满足前面任务时间的机器肯定满足后面时间的机器，只是看等级行不行     while(j<n&&machine[j].time>=task[i].time){//枚举每一个机器 若可以 还要选择等级最小的机器去完成   level[machine[j].level]++;  //将满足时间的机器的等级标记j++;   }for(int k=task[i].level;k<=100;k++){  //将任务的起始等级枚举  若有机器存在，则可以完成//因为之前加入进去满足时间的机器了 if(level[k]>0){level[k]--;re+=task[i].time*500+task[i].level*2;sum++; break;  //跳出 }}} printf("%d %lld\n",sum,re);}return 0;}