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 = ;
 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 = ;
     scanf("%d", &n);
     for(int i = ; i <= n; i++) {
         scanf("%d%d", &c[i].a, &c[i].b);
         c[i].num = i;
     }
     sort(c + , c + n + );
     Cow x;
     x.a = c[].b;
     x.b = ++top;
     ans[c[].num] = top;
     Q.push(x);
     for(int i = ; i <= n; i++) {
         //printf("i:%d c[i].b:%d Q.top().a:%d\n", 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\n", top);
     for(int i = ; i <= n; i++) {
         printf("%d\n", ans[i]);
     }
     return ;
 }

AC代码