2014 Multi-University Training Contest 1/HDU4864_Task(贪心)
解题报告
题意,有n个机器。m个任务。
每一个机器至多能完毕一个任务。对于每一个机器,有一个最大执行时间Ti和等级Li,对于每一个任务,也有一个执行时间Tj和等级Lj。仅仅有当Ti>=Tj且Li>=Lj的时候,机器i才干完毕任务j,并获得500*Tj+2*Lj金钱。
问最多能完毕几个任务,当出现多种情况时,输出获得金钱最多的情况。
对任务和机器进行从大到小排序,从最大时间且最大等级的任务開始,选取全部符合时间限制的机器中等级要求最低的,这样能保证以下任务可选择的机器最多,如果选择的机器中等级要求最大的,接下去有一任务同样时间。却可能找不到可用的机器。
如机器(200,5)(200,2)任务(100,2)(99,4)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 100010
#define LL __int64
using namespace std;
struct node
{
int t,l;
} ma[N],ta[N];
int n,m;
int cmp(node a,node b)
{
if(a.t==b.t)
return a.l>b.l;
else return a.t>b.t;
}
int main()
{
int i,j,k;
while(~scanf("%d%d",&n,&m))
{
LL sum=0;
int cnt=0;
for(i=0; i<n; i++)
scanf("%d%d",&ma[i].t,&ma[i].l);
for(i=0; i<m; i++)
scanf("%d%d",&ta[i].t,&ta[i].l);
sort(ma,ma+n,cmp);
sort(ta,ta+m,cmp);
int cc[110];
memset(cc,0,sizeof(cc));
for(i=0,j=0; i<m; i++)
{
while(j<n&&ta[i].t<=ma[j].t)
{
cc[ma[j].l]++;
j++;
}
for(k=ta[i].l; k<=100; k++)
{
if(cc[k])
{
sum+=(500*ta[i].t+2*ta[i].l);
cnt++;
cc[k]--;
break;
}
}
}
printf("%d %I64d\n",cnt,sum);
}
return 0;
}
Task
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1297 Accepted Submission(s): 321
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.
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.
1 2
100 3
100 2
100 1
1 50004
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