Task
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)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.
代码:
#include<bits/stdc++.h>
using namespace std;
struct point
{
int x,y;
};
point a[100000],b[100000];
int s[150];
bool cmp(point a,point b)
{
if(a.x==b.x)return a.y>b.y;
return a.x>b.x;
}
int main()
{
long long i,n,m,num,money;
while(cin>>n>>m)
{
for(i=1;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].y);
for(i=1;i<=m;i++)scanf("%d%d",&b[i].x,&b[i].y);
sort(a+1,a+1+n,cmp);
sort(b+1,b+1+m,cmp);
num=0;
money=0;
memset(s,0,sizeof(s));
int j=1;
for(int i=1;i<=m;i++)
{
while(j<=n&&a[j].x>=b[i].x)
{
s[a[j].y]++;
j++;
}
for(int k=b[i].y;k<=100;k++)if(s[k]>0)
{
num++;
money+=500*b[i].x+2*b[i].y;
s[k]--;
break;
}
}
printf("%d %lld\n",num,money);
}
return 0;
}