zoukankan      html  css  js  c++  java
  • hdu 4864 Task---2014 Multi-University Training Contest 1

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4864


    Task

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1035    Accepted Submission(s): 245


    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
     

    Recommend
    We have carefully selected several similar problems for you:  4871 4869 4868 4867 4866 
     


    这是一道贪心的题,要求任务数最多,并且钱也要最多。

    那么我们首先先对任务排序,先按时间排序,若时间一样,则依照level排。

    之后呢,我们依次选择时间和level最接近该任务的就可以。

    因为这道题数据范围比較大,在tle了n次之后,所以我參考了某神牛的博客,採用了set和二分查找进行优化,时间复杂度瞬间就降下来了。。。


    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cmath>
    #include<cstring>
    #include<cstdlib>
    #include<vector>
    #include<set>
    #include<map>
    #include<bitset>
    using namespace std;
    typedef long long ll;
    const int MAX=100005;
    struct Task{
        int x,y;
    }t[MAX];
    bool cmp(Task a,Task b){if(a.x==b.x) return a.y>b.y;else return a.x>b.x;};
    inline ll f(int x,int y){return 500*x+2*y;}
    int main(){
        int n,m;
        while(cin>>n>>m){
            multiset<int> st[105];
            int maxlevel=0,cnt=0;
            ll sum=0;
            for(int i=0;i<n;i++){
                int x,y;
                scanf("%d%d",&x,&y);
                st[y].insert(x);
                maxlevel=max(maxlevel,y);
            }
            for(int i=0;i<m;i++){scanf("%d%d",&t[i].x,&t[i].y);}
            sort(t,t+m,cmp);
            for(int i=0;i<m;i++){
                int u=t[i].x,v=t[i].y;
                for(int j=v;j<=maxlevel;j++){
                    if(st[j].empty()) continue;
                    multiset<int>::iterator it=st[j].lower_bound(u);
                    if(it==st[j].end()||*it<u) continue;
                    else{
                        cnt++;sum+=f(u,v);
                        st[j].erase(it);
                        break;
                    }
                }
            }
            cout<<cnt<<" "<<sum<<endl;
        }
        return 0;
    }
    



  • 相关阅读:
    Qt C++中的关键字explicit——防止隐式转换(也就是Java里的装箱),必须写清楚
    有栖川有栖《马来铁道之谜》读后感
    Qt多国语言QT_TR_NOOP和QT_TRANSLATE_NOOP
    Qt调用VC++生成的动态链接库
    QTabWidget添加自定义样式
    Qt跨线程信号和槽的连接(默认方式是直连和队列的折中)
    OO五大原则
    《Head First Python》学习笔记03 异常处理
    使用Qt实现MDI风格的主窗体
    Qt中文乱码问题(比较清楚,同一个二进制串被解释成不同的语言)
  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/4172779.html
Copyright © 2011-2022 走看看