Stall Reservations (POJ 3190 greedy)

Language:DefaultStall Reservations Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3394 Accepted: 1215 Special Judge Description Oh those picky n (1 <= n <= 50,000) cows! They is so picky, each one is only being milked over some precise time interval A. B (1 <= a <= b <= 1,000,000), which includes both times A and B. Obviously, FJ must create A reservation system t o determine which stall each cow can is assigned for her milking time. Of course, no cow would share such a private moment with other cows. Help FJ by determining: The minimum number of stalls required in the barn so, each cow can has her private milking period An assignment of cows to these stalls over time Many answers is correct for each test dataset; A program would grade your answer. Input Line 1: A single integer, N Lines 2..n+1:line i+1 describes cow i ' s milking interval with the space-separated integers. Output Line 1:the minimum number of stalls the barn must has. Lines 2..n+1:line i+1 describes the stall to which cow I'll be a 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 stalls is possible. Source Usaco 2006 February Silver

Test instructions: Given the time period for milking each cow, a machine can only work on a cow at a time, ask at least how many machines are needed, and output the machine number used by each cow.

Idea: First according to the starting time of each cow from small to large sort, maintain a priority queue, N cows in turn, the time to end the first out of the queue, compared to the earliest cattle end time and the current ready to join the cattle start time, if the former is smaller than the latter, then the current cattle can be used in front of empty machine, No one wants to add a new machine.

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set > #include <queue> #pragma comment (linker, "/stack:102400000,102400000") #define MAXN 50010#define MAXN 2005# Define mod 1000000009#define INF 0x3f3f3f3f#define pi ACOs ( -1.0) #define EPS 1e-6#define Lson rt<<1,l,mid#define RSO N rt<<1|1,mid+1,r#define FRE (i,a,b) for (i = A, I <= b; i++) #define FRL (i,a,b) for (i = A; I < b; i++) #define Mem (T, v) memset ((t), V, sizeof (t)) #define SF (n) scanf ("%d", &n) #define SFF (A, b) scanf ("%d%d", &a, & AMP;B) #define SFFF (a,b,c) scanf ("%d%d%d", &a, &b, &c) #define PF printf#define DBG pf ("hi\n"    ) typedef long Long ll;using namespace std;struct cow{int s,t; int local;}    Cow[maxn];int cmp (Cow A,cow b) {if (A.S==B.S) return a.t<b.t; return A.S&LT;B.S;} PrioRity_queue<cow>q;int ans[maxn];int n;bool operator< (Cow A,cow b) {return a.t>b.t;}    int main () {int i,j;    Cow St;            while (~SF (n)) {FRL (i,0,n) {SFF (cow[i].s,cow[i].t);        Cow[i].local=i;        } sort (cow,cow+n,cmp);        Q.push (Cow[0]);        int cnt=1;        ans[cow[0].local]=cnt; FRL (i,1,n) {if (!                Q.empty () &&q.top (). T<cow[i].s) {st=q.top ();                Q.pop ();            ans[cow[i].local]=ans[st.local];            } else ans[cow[i].local]=++cnt;        Q.push (Cow[i]);        } printf ("%d\n", CNT);    FRL (i,0,n) pf ("%d\n", Ans[i]); } return 0;} /*51 102 43 65 84 7*/

Stall Reservations (POJ 3190 greedy)