HDU 4864 Task(貪心)

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TaskTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3587 Accepted Submission(s): 930


Problem DescriptionToday 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.
InputThe 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.
OutputFor 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

AuthorFZU
Source2014 Multi-University Training Contest 1方法一：機器匹配任務：首先機器和任務的時間和層級都按  從小到大排，再從第一個機器開始，找出從第一個任務開始找出所有任務的時間小於等於這台機器的時間，用數組計下每個等級出現的次數。再從等級數組中找一個等級最接近這台機器的等級。這樣再找任務的時間複雜度是線性為為O(m)。
方法二：任務匹配機器：首先機器和任務的時間和層級都按  從大到小排，再從第一個任務開始，找出從第一個機器開始找出所有機器的時間大於等於這個任務的時間，用數組計下每個等級出現的次數。再從等級數組中找一個等級最接近這個任務的等級。這樣再找機器的時間複雜度是線性為為O(n)。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{    int x,y;};node mach[100005],task[100005];bool cmp(node a,node b){    if(a.x==b.x)    return a.y>b.y;    return a.x>b.x;}int main(){    int n,m,ans,x,y,level[105];    __int64 sum;    while(scanf("%d%d",&n,&m)>0)    {        memset(level,0,sizeof(level));        for(int i=0;i<n;i++)            scanf("%d%d",&mach[i].x,&mach[i].y);        for(int i=0;i<m;i++)        scanf("%d%d",&task[i].x,&task[i].y);        sort(mach,mach+n,cmp);        sort(task,task+m,cmp);        ans=0; sum=0;        int j=0;        for(int i=0;i<m;i++)        {            while(j<n&&mach[j].x>=task[i].x)level[mach[j++].y]++;            for(int lev=task[i].y; lev<=100;lev++)            if(level[lev])            {                ans++;                 sum+=500*task[i].x+2*task[i].y;                 level[lev]--;                break;            }        }        printf("%d %I64d\n",ans,sum);    }}

HDU 4864 Task(貪心)