每日一题 day37 打卡
Analysis
经典的带限期和罚款的单位时间任务调度问题
将 val 从大到小排序,优先处理罚款多的,将任务尽量安排在期限之前,并且靠后,如果找不到,则放在最后面
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
#define maxn 500+10
#define rep(i,s,e) for(register int i=s;i<=e;++i)
#define dwn(i,s,e) for(register int i=s;i>=e;--i)
using namespace std;
inline int read()
{
int x=;
bool f=;
char c=getchar();
for(; !isdigit(c); c=getchar()) if(c=='-') f=;
for(; isdigit(c); c=getchar()) x=(x<<)+(x<<)+c-'';
if(f) return x;
return -x;
}
inline void write(int x)
{
if(x<){putchar('-');x=-x;}
if(x>)write(x/);
putchar(x%+'');
}
int m,n,sum;
struct node
{
int tim,_val;
}x[maxn];
int book[maxn];
bool cmp(node x,node y)
{
return x._val>y._val;
}
signed main()
{
m=read();n=read();
rep(i,,n) x[i].tim=read();
rep(i,,n) x[i]._val=read();
sort(x+,x+n+,cmp);
rep(i,,n)
{
int flag=;
dwn(j,x[i].tim,)
if(book[j]==)
{
book[j]=;
flag=;
break;
}
if(flag==)
{
dwn(j,n,)
if(book[j]==)
{
book[j]=;
break;
}
sum+=x[i]._val;
}
}
write(m-sum);
return ;
}
请各位大佬斧正(反正我不认识斧正是什么意思)