poj Stall Reservations(贪心)

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 period
An 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.

#include<cstdio>#include<algorithm>#include<cstring>#include<queue>using namespace std;const int M = 5e4 + 10;struct Node{    int l,r,id;    bool operator < (const Node &a)const    {        if(r==a.r)            return l > a.l;        return r > a.r;    }}cow[M];priority_queue<Node> q;int adr[M];bool cmp(Node a, Node b){    if(a.l==b.l)        return a.r < b.r;    return a.l < b.l;}int main(){    int n, t1, t2;    scanf("%d", &n);    for(int i=1;i<=n;i++)    {        scanf("%d%d", &cow[i].l, &cow[i].r);        cow[i].id = i;    }    sort(cow+1, cow+n+1, cmp);    int tail = 1;    q.push(cow[1]);    adr[cow[1].id] = tail;    for(int i=2;i<=n;i++)    {        if(!q.empty()&&cow[i].l>q.top().r)        {            adr[cow[i].id] = adr[q.top().id];            q.pop();        }        else            adr[cow[i].id] = ++tail;        q.push(cow[i]);    }    printf("%d\n", tail);    for(int i=1;i<=n;i++)        printf("%d\n", adr[i]);    return 0;}