题目链接
题解
以前看别人博客，在考场上用费用流做，一直以为这题是毒瘤网络流题
没想到竟然是贪心模拟题。。。
如果只有一个蔬菜呢？这就是一个经典的普及难度的贪心，正着推面临优先选择的困难，而逆着推由于不存在淘汰，所以可以贪心选最大的
首先(s_i)的限制很容易处理，只需将每一个蔬菜分出一个价值(a_i + s_i)且过期时间为该蔬菜最后一个的蔬菜
现在我们计算出每个蔬菜最晚放置的时间点，将每一天看做一个盒子，我们贪心地优先将价值大的蔬菜从它能放入的地方一直往前放
由于每个盒子最多放(10)个，我们用并查集合并相邻的放满的盒子，全部放置满只需(O(10^5m))
但是这样对于每一个(p)都要算一次，实则不然，我们先算出最大的(p)的答案，发现前(p - 1)天能使用的蔬菜包含前(p)天能使用的蔬菜，我们用堆维护已放入的蔬菜，对于前(p - 1)天，我们只需把堆中的蔬菜减至((p - 1)m)即可
复杂度(O(10^5mlogn))
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 200005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
priority_queue<LL,vector<LL>,greater<LL> > q;
LL cost[maxn],los[maxn],tot[maxn],E[maxn];
LL p[maxn],Pi = 100000,used[maxn],ans;
int n,m,K,N,pre[maxn],id[maxn];
inline int find(int u){return u == pre[u] ? u : pre[u] = find(pre[u]);}
inline bool cmp(const int& a,const int& b){
return cost[a] > cost[b];
}
void work(){
REP(i,N) id[i] = i;
REP(i,Pi) pre[i] = i; used[0] = m;
sort(id + 1,id + 1 + N,cmp);
LL sum,Out;
for (int i = 1; i <= N; i++){
int u = id[i],now;
if (u > n){
now = find(E[u]);
if (!now) continue;
used[now]++;
ans += cost[u];
q.push(cost[u]);
if (used[now] == m) pre[now] = now - 1;
}
else {
Out = 0; now = E[u];
for (; ; now--){
now = find(now);
if (!now) break;
sum = tot[u] - los[u] * (now - 1) - Out;
if (!sum && !los[u]) break;
else if (!sum) continue;
if (sum >= m - used[now]){
Out += m - used[now];
ans += cost[u] * (m - used[now]);
pre[now] = now - 1;
REP(j,m - used[now]) q.push(cost[u]);
used[now] = m;
}
else {
Out += sum;
used[now] += sum;
ans += cost[u] * sum;
REP(j,sum) q.push(cost[u]);
}
}
}
}
p[Pi] = ans;
for (int i = Pi - 1; i; i--){
while (q.size() > 1ll * m * i) ans -= q.top(),q.pop();
p[i] = ans;
}
}
int main(){
//freopen("1.in","r",stdin);
//freopen("1.out","w",stdout);
N = n = read(); m = read(); K = read(); int s;
for (int i = 1; i <= n; i++){
cost[i] = read(); s = read(); tot[i] = read(); los[i] = read();
cost[++N] = cost[i] + s; tot[N] = 1; tot[i]--;
if (!los[i]) E[i] = E[N] = Pi;
else if (tot[i] / los[i] >= Pi) E[i] = E[N] = Pi;
else {
E[i] = tot[i] / los[i] + (tot[i] % los[i] != 0);
if (E[i] * los[i] > tot[i]) E[N] = E[i];
else E[N] = E[i] + 1;
}
}
work();
while (K--) printf("%lld
",p[read()]);
return 0;
}