JZOJ 4863. 【GDOI2017模拟11.5】Market

题目

题解

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define N 100010
#define LL long long
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
using namespace std;
struct note
{
LL c,v,t;
};note a
;
struct note1
{
LL T,M,w;
};note1 b
;
LL i,j,k,l,r,mid,n,m,lm,ans
,_2[22],f
;
bool v
;
bool cmp(note x,note y){return x.t<y.t;}
bool cmp1(note1 x,note1 y){return x.T<y.T;}
int main()
{
freopen("market.in","r",stdin);
freopen("market.out","w",stdout);
_2[1]=1;
fo(i,2,20) _2[i]=_2[i-1]*2;
scanf("%lld%lld",&n,&m);
fo(i,1,n)
{
scanf("%lld%lld%lld",&a[i].c,&a[i].v,&a[i].t);
lm=max(a[i].v,lm);
}
lm*=n;
fo(i,1,m)
{
scanf("%lld%lld",&b[i].T,&b[i].M);
b[i].w=i;
}
sort(a+1,a+n+1,cmp);
sort(b+1,b+m+1,cmp1);
memset(f,127,sizeof(f));
f[0]=0;
j=1;
fo(i,1,m)
{
while (j<=n && a[j].t<=b[i].T)
{
fd(k,lm,a[j].v)
f[k]=min(f[k],f[k-a[j].v]+a[j].c);
fd(k,lm-1,1)f[k]=min(f[k+1],f[k]);
j++;
}
LL temp;
l=0;r=lm;
while (l<=r)
{
mid=(l+r)/2;
if (f[mid]<=b[i].M) temp=mid,l=mid+1;
else r=mid-1;
}
ans[b[i].w]=temp;
}
fo(i,1,m) printf("%lld\n",ans[i]);
}