poj3190 Stall Reservations

我一开始想用线段树，但是发现还要记录每头cow所在的棚......

无奈之下选择正解：贪心。

用priority_queue来维护所有牛棚中结束时间最早的那个牛棚，即可得出答案。

注意代码实现的细节。

#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 50010;
struct Cow {
    int a, b, num;
    bool operator < (const Cow& x) const {
        return this->a < x.a;
    }
    bool operator > (const Cow& x) const {
        return this->a > x.a;
    }
}c[N];
priority_queue< Cow, vector<Cow>, greater<Cow> >Q;
int ans[N];
int main() {
    int n, top = 0;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d%d", &c[i].a, &c[i].b);
        c[i].num = i;
    }
    sort(c + 1, c + n + 1);
    Cow x;
    x.a = c[1].b;
    x.b = ++top;
    ans[c[1].num] = top;
    Q.push(x);
    for(int i = 2; i <= n; i++) {
        //printf("i:%d c[i].b:%d Q.top().a:%d
", c[i].num, c[i].b, Q.top().a);
        if(Q.top().a < c[i].a) {
            x = Q.top();
            Q.pop();
            x.a = c[i].b;
            ans[c[i].num] = x.b;
            Q.push(x);
        }
        else {
            x.a = c[i].b;
            x.b = ++top;
            ans[c[i].num] = top;
            Q.push(x);
        }
    }
    printf("%d
", top);
    for(int i = 1; i <= n; i++) {
        printf("%d
", ans[i]);
    }
    return 0;
}