Description
http://www.lydsy.com/JudgeOnline/upload/Noi2017D2.pdf
Solution
网上大部分都是并查集写法，但是有大神写了非并查集写法，特别容易理解
首先 (s_i) 的限制，只需将每一个蔬菜分出一个价值为 (a_i+s_i) 且过期时间为该蔬菜最晚的一天的蔬菜
把时间倒序之后，问题转化为每个蔬菜会在第几天出现，每天贪心选择价值最大的即可
先求出 (max {p_i }) 的答案，然后递推 ([1,max {pi}-1]) 的答案：每次删除价值最小的 (m) 单位蔬菜，不难发现所有蔬菜的销售时间依然合法
#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=100000+10,P=100000;
int n,m,k,a[MAXN],c[MAXN],x[MAXN],s[MAXN],vis[MAXN];
ll ans[MAXN],sale[MAXN],tot;
struct node{
ll id,val;
inline bool operator < (const node &A) const {
return val<A.val;
};
inline bool operator > (const node &A) const {
return val>A.val;
};
};
std::vector<int> V[MAXN];
std::queue<node> red;
std::priority_queue<node> q1;
std::priority_queue< node,std::vector<node>,std::greater<node> > q2;
template<typename T> inline void read(T &x)
{
T data=0,w=1;
char ch=0;
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
x=data*w;
}
template<typename T> inline void write(T x,char ch=' ')
{
if(x<0)putchar('-'),x=-x;
if(x>9)write(x/10);
putchar(x%10+'0');
if(ch!=' ')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
int main()
{
read(n);read(m);read(k);
for(register int i=1;i<=n;++i)
{
read(a[i]),read(s[i]),read(c[i]),read(x[i]);
if(x[i]==0)V[P].push_back(i);
else V[min(P,(c[i]+x[i]-1)/x[i])].push_back(i);
}
for(register int i=P;i>=1;--i)
{
for(register int j=0,lt=V[i].size();j<lt;++j)q1.push({V[i][j],a[V[i][j]]+s[V[i][j]]});
ll rest=m;
while(!q1.empty()&&rest)
{
node now=q1.top();
q1.pop();
if(vis[now.id])
{
ll res=min(rest,c[now.id]-(i-1)*x[now.id]-sale[now.id]);
ans[P]+=1ll*res*a[now.id];sale[now.id]+=res;rest-=res;
if(sale[now.id]<c[now.id])red.push(now);
}
else
{
vis[now.id]=1;
ans[P]+=now.val;sale[now.id]++;rest--;
if(sale[now.id]<c[now.id])q1.push((node){now.id,a[now.id]});
}
}
while(!red.empty())q1.push(red.front()),red.pop();
}
for(register int i=1;i<=n;tot+=sale[i],++i)
if(sale[i]==1)q2.push((node){i,a[i]+s[i]});
else if(sale[i])q2.push((node){i,a[i]});
for(register int i=P-1;i>=1;--i)
{
ans[i]=ans[i+1];
ll rest=tot-1ll*m*i;
if(rest<=0)continue;
else
while(!q2.empty()&&rest)
{
node now=q2.top();
q2.pop();
if(sale[now.id]==1)sale[now.id]--,rest--,ans[i]-=now.val;
else
{
int res=min(rest,sale[now.id]-1);
sale[now.id]-=res,rest-=res,ans[i]-=1ll*res*now.val;
if(sale[now.id]==1)q2.push((node){now.id,a[now.id]+s[now.id]});
else if(sale[now.id])q2.push((node){now.id,a[now.id]});
}
}
tot=1ll*m*i;
}
for(register int i=1,x;i<=k;++i)read(x),write(ans[x],'
');
return 0;
}