poj 3190 Stall Reservations（区间贪心，优先队列）

Stall Reservations Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5390 Accepted: 1945 Special Judge

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 A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, 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 stalls required in the barn so that each cow can have her private milking periodAn assignment of cows to these stalls over time Many 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 stalls the 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

5 1 10 2 4 3 6 5 8 4 7

Sample Output

4 1 2 3 2 4

Hint

Explanation of the sample:  Here's a graphical schedule for this output:  Time 1 2 3 4 5 6 7 8 9 10 Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>> Stall 2 .. c2>>>>>> c4>>>>>>>>> .. .. Stall 3 .. .. c3>>>>>>>>> .. .. .. .. Stall 4 .. .. .. c5>>>>>>>>> .. .. .. Other outputs using the same number of stalls are possible.

Source

USACO 2006 February Silver

题意：每头牛独享一个牛栏里的[A, B]，问最少需要几个牛栏。

思路：优先选A最小的，维持一个牛栏B最小堆，将新来的奶牛塞进B最小的牛栏里。

代码：

#include<iostream> #include<queue> #include<cstdio> #include<algorithm> using namespace std; const int maxn = 50005; struct node { int x, y, id; bool operator <(const node a) const { return y > a.y; } }a[maxn]; int ans[maxn]; bool cmp(node a, node b) { if(a.x == b.x) return a.y < b.y; else return a.x < b.x; } int main(void) { int n; while(~scanf("%d", &n)) { int k = 1; priority_queue<node> pq; for(int i = 0; i < n; i++) { scanf("%d%d", &a[i].x, &a[i].y); a[i].id = i; } sort(a, a+n, cmp); pq.push(a[0]); ans[a[0].id] = k++; for(int i = 1; i < n; i++) { node t = pq.top(); if(t.y < a[i].x) { ans[a[i].id] = ans[t.id]; pq.pop(); pq.push(a[i]); } else { ans[a[i].id] = k++; pq.push(a[i]); } } printf("%d\n", k-1); for(int i = 0; i < n; i++) printf("%d\n", ans[i]); } return 0; }