Bzoj2590 [usaco2012 Feb] cow coupons

Okay... I thought about it for a long time... Although I know it is greedy...

Each time we look for the two types of cattle that have not been bought with the minimum price, compare a = the largest current price difference + the lowest price of the current coupon, and B = the lowest price of the current non-coupon

So... We want

First, maintain two small-rooted heaps, indicating the price of cattle bought with coupons and the price of cattle bought without coupons.

There is also a big root heap called recover, which indicates the price difference between the current several cows using coupons (the price difference is defined as the difference between the non-preferential price and the preferential price)

A and B can be bought if they are small...

 

 1 /************************************************************** 2     Problem: 2590 3     User: rausen 4     Language: C++ 5     Result: Accepted 6     Time:140 ms 7     Memory:3332 kb 8 ****************************************************************/ 9  10 #include <cstdio>11 #include <algorithm>12 #include <queue>13  14 using namespace std;15 typedef long long ll;16 const int N = 50005;17 struct data{18     int w, v;19     data(void){}20     data(int x, int y) : w(x), v(y) {}21 };22 inline bool operator < (const data a, const data b){23     return a.v > b.v;24 }25 int cnt, n, k;26 int p[N], c[N];27 bool vis[N];28 ll m;29 priority_queue <data> h, H;30 priority_queue <ll, vector<ll>, greater<ll> > Re;31  32 inline int read(){33     int x = 0;34     char ch = getchar();35     while (ch < ‘0‘ || ch > ‘9‘)36         ch = getchar();37     while (ch >= ‘0‘ && ch <= ‘9‘){38         x = x * 10 + ch - ‘0‘;39         ch = getchar();40     }41     return x;42 }43  44 int main(){45     int i;46     ll cost;47     data T;48     n = read(), k = read();49     scanf("%lld", &m);50     for (i = 1; i <= n; ++i){51         p[i] = read(), c[i] = read();52         h.push(data(i, c[i]));53         H.push(data(i, p[i]));54     }55     while (!Re.empty()) Re.pop();56     for (i = 1; i <= k; ++i)57         Re.push(0);58     while (m > 0 && cnt < n){59         while (vis[h.top().w])60             h.pop();61         while (vis[H.top().w])62             H.pop();63         if (Re.top() + h.top().v < H.top().v){64             T = h.top(), cost = Re.top() + T.v;65             if (m < cost) break;66             m -= cost;67             Re.pop();68             Re.push(p[T.w] - c[T.w]);69             vis[T.w] = 1;70         }else{71             T = H.top(), cost = T.v;72             if (m < cost) break;73             m -= cost;74             vis[T.w] = 1;75         }76         ++cnt;77     }78     printf("%d\n", cnt);79     return 0;80 }

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Bzoj2590 [usaco2012 Feb] cow coupons