poj3190 Stall Reservations

思路：

关于区间的贪心。

先对区间排序。再从前到后扫描，用堆维护当前最小的区间右端点right和对应的stall编号id，如果当前处理的区间的左端点大于right，则可以把这个区间放在编号为id的stall后面；否则重新开一个stall。

实现：

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
using namespace std;

typedef pair<int, int> P;
struct node
{
    int index, left, right;
};
node a[50005];
int cow[50005];
int n;
priority_queue<P, vector<P>, greater<P> > pq;

bool cmp(const node & a, const node & b)
{
    if (a.left != b.left)
    {
        return a.left < b.left;
    }
    return a.right < b.right;
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        a[i].index = i;
        scanf("%d %d", &a[i].left, &a[i].right);
    }
    sort(a + 1, a + n + 1, cmp);
    int cnt = 0;
    pq.push(P(a[1].right, ++cnt));
    cow[a[1].index] = cnt;
    for (int i = 2; i <= n; i++)
    {
        if (a[i].left > pq.top().first)
        {
            P tmp = pq.top();
            pq.pop();
            cow[a[i].index] = tmp.second;
            P tmp2(a[i].right, tmp.second);
            pq.push(tmp2);
        }
        else
        {
            pq.push(P(a[i].right, ++cnt));
            cow[a[i].index] = cnt;
        }
    }
    cout << cnt << endl;
    for (int i = 1; i <= n; i++)
    {
        cout << cow[i] << endl;
    }
    return 0;
}