hdoj Task
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Task
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1369 Accepted Submission(s): 341
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 2100 3100 2100 1
1 50004
题目链接点击打开链接
本题的策略是贪心,
对于价值c=500*xi+2*yi,yi最大影响100*2<500,所以就是求xi总和最大。可以先对机器和任务的时间从大到小排序。从最大时间的任务开始,找出满足任务时间要求的所有机器,从中找出等级最低且满足任务等级要求的机器匹配。依次对任务寻找满足要求的机器
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct s{
int x,y;
}a[100010],b[100010];
int cmp(const void *a,const void *b)
{
if((*(struct s *)a).x==(*(struct s *)b).x)
return (*(struct s *)b).y-(*(struct s *)a).y;
else
return (*(struct s *)b).x-(*(struct s *)a).x;
}
int main()
{
int m,n,d[110];
while(scanf("%d%d",&n,&m)!=EOF)
{
int i,j=0,k;
__int64 c=0,sum=0;
memset(d,0,sizeof(d));
//memset(v,0,sizeof(v));
for(i=0;i<n;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
//a[i].v=0;
}
for(i=0;i<m;i++)
{
scanf("%d%d",&b[i].x,&b[i].y);
//b[i].v=0;
}
qsort(a,n,sizeof(a[0]),cmp);
qsort(b,m,sizeof(b[0]),cmp);
/*for(i=0;i<m;i++)
printf("%d %d ",b[i].x,b[i].y);*/
for(i=0;i<m;i++)
{
while(j<n&&a[j].x>=b[i].x)
{
d[a[j].y]++;//统计等级数
j++;//j无需从0开始遍历,因为已经统计好等级数,且是按时间排的,比上一个任务时间大,必定比现在一个任务时间大
}
for(k=b[i].y;k<=100;k++)
{
if(d[k])
{
sum+=500*b[i].x+2*b[i].y;
c++;
d[k]--;
//v[j]=0;
break;
}
}
}
printf("%I64d %I64d\n",c,sum);
}
}
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