Sicily 13983. Milk Scheduling

Source: Internet
Author: User

13983. Milk Scheduling Constraints

Time limit:1 secs, Memory limit:256 MB

Description

Farmer John has N cows this need to being milked (1 <= N <=), each of the which takes only one unit of time to milk.

Being impatient animals, some cows would refuse to being milked if Farmer John waits too long to milk them. More specifically, cow I produces g_i gallons of milk (1 <= g_i <=), but only if she's milked before a Deadlin E at time d_i (1 <= d_i <= 10,000). Time starts at t=0, so at the most x total cows can is milked prior to a deadline at time T=x.

Farmer John determine the maximum amount of milk that he can obtain if he milks the cows optimally.

Input

* Line 1:the value of N.

* Lines 2..1+n:line i+1 contains the integers g_i and d_i.

Output

* Line 1:the Maximum number of gallons of milk Farmer John can obtain.

Sample Input
410 37 58) 12 1
Sample Output
25
Hint

In the sample, there is 4 cows. The first produces gallons of milk if milked by time 3, and so on. Farmer John Milks Cow 3 first, giving up on cow 4 since she cannot being milked by hers deadline due to the conflict with cow 3. Farmer John then milks cows 1 and 2.

Problem Source

2015 every Monday match sixth game

According to G Big first, D small priority, for a cow, priority to the point of time as close as possible to D, so as far as possible delay, give the other cows full time ...

#include <iostream> #include <algorithm>using namespace std;struct cow {int g, d;cow (int g = 0, int d = 0) {thi S->g = G;this->d = D;}}; BOOL CMP (const COW & C1, Const Cow & C2) {if (c1.g = = c2.g) return c1.d < C2.d;else return c1.g > c2.g;} Cow c[10005];bool used[10005];int havetime (int d) {for (int i = D; i > 0; i--) if (!used[i]) return I;return 0;} int main () {Std::ios::sync_with_stdio (false), int n;cin >> n;for (int i = 0; i < n; i++) cin >> C[I].G >& Gt C[i].d;sort (c, C + N, CMP); int t = 4, ans = 0;for (int i = 0; i < n; i++) {int feedday = Havetime (C[I].D); if (Feedday) {Used[feedday] = True;ans + = C[I].G;}} cout << ans << endl;return 0;}

Sicily 13983. Milk Scheduling

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