LA 4328 Priest John's Busiest Day (Greedy)

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2329

　　很好的一道贪心题。在贪心的时候，因为时间安排是要连续的，所以用区间的中点作为排序的依据。中点靠前的区间要先贪心，标记当前最后的时间点。

View Code

 1 #define REP(i, n) for (int i = 0; i < (n); i++)
 2 
 3 struct Cerem {
 4     int begin, end;
 5     int len, mid;
 6     Cerem(int b = 0, int e = 0) {
 7         begin = b, end = e;
 8         len = (e - b >> 1) + 1;
 9         mid = b + len;
10     }
11 } cerem[N];
12 
13 bool cmp(Cerem a, Cerem b) {
14     return a.mid < b.mid;
15 }
16 
17 void input(int n) {
18     int b, e;
19     REP(i, n) {
20         scanf("%d%d", &b, &e);
21         cerem[i] = Cerem(b, e);
22     }
23     sort(cerem, cerem + n, cmp);
24 }
25 
26 bool work(int n) {
27     int end = 0;
28     REP(i, n) {
29         if (end <= cerem[i].begin) end = cerem[i].begin + cerem[i].len;
30         else if (end + cerem[i].len <= cerem[i].end) end += cerem[i].len;
31         else return false;
32     }
33     return true;
34 }
35 
36 int main() {
37     int n;
38     while (~scanf("%d", &n) && n) {
39         input(n);
40         work(n) ? puts("YES") : puts("NO");
41     }
42     return 0;
43 }

——written by Lyon