2018 Multi-University Training Contest 8 - From ICPC to ACM

贪心 + 模拟

预处理出最便宜的原材料价格，然后用map储存每种价格的电脑数量。

这里因为每个月价格都在变化，所以可以用相对价格，也就是减去一个sigema(e[i])。

每次选择最小的卖，统计贡献，最后如果超出库存，反着删除即可（相当于没有生产这些电脑）

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline ll read(){
    ll ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
    return w ? -ret : ret;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template <typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template <typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}

const int N = 50005;
ll _, k, c[N], d[N], m[N], p[N], e[N], R[N], E[N];
map<ll, ll> record;
int main(){

    for(_ = read(); _; _ --){
        k = read(), record.clear();
        for(int i = 1; i <= k; i ++){
            c[i] = read(), d[i] = read(), m[i] = read(), p[i] = read();
        }
        for(int i = 1; i < k; i ++){
            e[i] = read(), R[i] = read(), E[i] = read();
        }
        for(int i = 1; i < k; i ++){
            if(c[i] + R[i] < c[i + 1]) c[i + 1] = c[i] + R[i];
        }
        ll ans = 0;
        int sum = 0, tot = 0;
        for(int i = 1; i <= k; i ++){
            record[c[i] + m[i] - sum] += p[i], tot += p[i];
            while(!record.empty() && d[i]){
                if(record.begin()->second > d[i]){
                    tot -= d[i];
                    record.begin()->second -= d[i];
                    ans += (record.begin()->first + sum) * d[i];
                    d[i] = 0;
                }
                else{
                    tot -= record.begin()->second;
                    d[i] -= record.begin()->second;
                    ans += (record.begin()->first + sum) * record.begin()->second;
                    record.erase(record.begin()->first);
                }
            }
            if(d[i]){
                ans = -1;
                break;
            }
            if(tot > e[i]){
                while(!record.empty() && tot > e[i]){
                    if(tot - record.rbegin()->second >= e[i]){
                        tot -= record.rbegin()->second;
                        record.erase(record.rbegin()->first);
                    }
                    else{
                        record.rbegin()->second -= (tot - e[i]);
                        tot = e[i];
                    }
                }
            }
            sum += E[i];
        }
        printf("%lld\n", ans);
    }
    return 0;
}