poj 3190 Stall Reservations

Stall Reservations Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6163   Accepted: 2250   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 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 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.

每一只奶牛要求在时间区间[A, B]内独享一个牛栏。问最少需要多少个牛栏。

贪心策略是优先满足A最小的奶牛，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。

问题的麻烦点在于如何求出每个区间是第几个牛栏。其实只需要保存每个区间是第几个，再保存放在是第几个牛栏。最后再用一次排序就可以了。

#include 
#include 
#include 
#include 
#include 
using namespace std;

const int maxn = 1e6 + 4;
struct node{
    int a, b, c, d;
};
node no[maxn];

bool cmp(node x, node y) {
    if(x.a == y.a) {
        return x.b < y.b;
    } else {
        return x.a < y.a;
    }
}
bool cmp2(node x, node y) {
    return x.c < y.c;
}

struct node2 {
    int x, y;
    bool operator < (const node2 &a) const {
        return x>a.x;//最小值优先
    }
};


int main()
{
    int n;
    priority_queue que;//最小值优先;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i) {
        scanf("%d%d", &no[i].a, &no[i].b);
        no[i].c = i;
    }
    sort(no, no + n, cmp);
    que.push(node2{no[0].b, 1});
    int ans = 1;
    no[0].d = 1;
    for(int i = 1; i < n; ++i) {
        node2 temp = que.top();
        int tempx = temp.x;
        int tempy = temp.y;
        if(no[i].a > tempx) {
            que.pop();
            que.push(node2{no[i].b, tempy});
            no[i].d = tempy;
        } else {
            ans++;
            que.push(node2{no[i].b, ans});
            no[i].d = ans;
        }
    }
    sort(no, no + n, cmp2);
    printf("%d\n", ans);
    for(int i = 0; i < n; ++i) {
        printf("%d\n", no[i].d);
    }
    return 0;
}