Poj 3190 stall reservations (Greedy + priority queue optimization)

Description Oh those picky N (1 <=n <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval .. B (1 <= A <= B <= 1,000,000), where des both times A and B. obviusly, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. of course, no cow will share such a private moment with other cows. Help FJ by determining: The minimum number of stils required in the barn so that each cow can have her private milking Period An assignment of cows to these stils over time Specified answers are correct for each test dataset; a program will grade your answer. Input Line 1: A single integer, n Lines 2. n + 1: line I + 1 describes cow I's milking interval with two space-separated integers. Output Line 1: the minimum number of stallthe barn must have. Lines 2. n + 1: line I + 1 describes the stall to which cow I will be assigned for her milking period. Sample Input 51 102 43 65 84 7 Sample output 412324 Hint Explanation of the sample: Here's a graphical schedule for this output: Time 1 2 3 4 5 6 7 8 9 10Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..Stall 3 .. .. c3>>>>>>>>> .. .. .. ..Stall 4 .. .. .. c5>>>>>>>>> .. .. .. Other outputs using the same number of stils are possible. Source Usaco 2006 February silver

Amount, greedy question =. =, Priority queue in STL

Greedy strategy: sorts the milking start time from big to small, so that the milking start time is not considered before;

Sort the elements in the queue by the milking end time from small to large, so that the first time can be determined without adding a slot =. =

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <limits. h ># include <queue> using namespace STD; const int maxn = 1000000 + 10; struct node {int St, Ed; int Pos; // The number of cows whose POS records are bool operator <(const node & A) const {If (ED =. ed) return st>. st; return ed>. ed ;}} A [maxn]; int used [maxn]; // used records the position where the cows are milking =. = Int CMP (node L1, node l2) {If (l1.st = l2.st) return l1.ed <l2.ed; return l1.st <l2.st;} int main () {int N; while (~ Scanf ("% d", & N) {priority_queue <node> q; For (INT I = 1; I <= N; I ++) {scanf ("% d", & A [I]. st, & A [I]. ed); A [I]. pos = I;} Sort (a + 1, A + n + 1, CMP); q. push (A [1]); int ans = 1; used [A [1]. pos] = 1; for (INT I = 2; I <= N; I ++) {If (! Q. empty () & Q. top (). ed <A [I]. st) // determine whether the condition is met {used [A [I]. pos] = used [q. top (). pos]; q. pop ();} else // The number of non-conforming slots plus one. At the same time, this cow should be added in the slot {ans ++; used [A [I]. pos] = ans;} Q. push (A [I]);} printf ("% d \ n", ANS); For (INT I = 1; I <= N; I ++) printf ("% d \ n", used [I]);} return 0 ;}