HDU 4864 task

Link: HDU 4864 task

Task Time Limit: 4000/2000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 843 accepted submission (s): 185


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

 
Source2014 multi-university training contest 1
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Question:

There are n machines and M tasks. Each machine can complete one task at most. For each machine, there is a maximum run time XI and level Yi. For each task, there is also a run time XJ and level YJ. Only when xi> = XJ and Yi> = YJ, machine I can complete task J and get 500 * XJ + 2 * YJ money. Ask how many tasks can be completed, and when there are multiple situations, output the most money.

Analysis:

When I did not make such a question during the competition, I could only say that I am too watery. I learned the basic idea of greed from the beginning. That is, first sort the tasks and machines in descending order (first time, time is the same and then by difficulty), and then use greedy solutions. At that time, the brain hole was too big. First, select a machine and select the task with the maximum profit that the machine can accomplish. However, when selecting the next machine, I forgot to consider the tasks ignored by the previous machine, in the end, this problem occurs when the infinite Wa is caused. The online question is to select a task and then select the appropriate minimum machine. I modified my code a little according to the question and AC it. In fact, it is the same as the idea of selecting a machine first. Alas, it's too stupid to do it.

Code:

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;#define maxn 100010struct node{    int tm, dif;}MCH[maxn], TSK[maxn];bool cmp(node a, node b){    if(a.tm==b.tm) return a.dif > b.dif;    return a.tm > b.tm;}int main(){    int n, m;    while(~scanf("%d%d", &n, &m))    {        for(int i = 1; i <= n; i++)            scanf("%d%d", &MCH[i].tm, &MCH[i].dif);        for(int i = 1; i <= m; i++)            scanf("%d%d", &TSK[i].tm, &TSK[i].dif);        sort(MCH+1, MCH+n+1, cmp);        sort(TSK+1, TSK+m+1, cmp);        int cnt = 0, c[102];        __int64 sum = 0;        memset(c, 0, sizeof(c));        for(int i = 1, j = 1; i <= m; i++)        {            while(j <= n && MCH[j].tm >= TSK[i].tm)            {                c[MCH[j].dif]++;                j++;            }            for(int k = TSK[i].dif; k <= 100; k++)            {                if(c[k])                {                    cnt++;                    c[k]--;                    sum += 500*TSK[i].tm+2*TSK[i].dif;                    break;                }            }        }        printf("%d %I64d\n", cnt, sum);    }    return 0;}

HDU 4864 task