BZOJ1727 [Usaco2006 Open]The Milk Queue 挤奶队列

贪心。。。我怎么不会QAQ【捂脸熊】

对于1、2两头牛，如果1号牛要排在2号牛前面才能时间更少，则

$$max(A_1 + B_1 + B_2, \ A_1 + A_2 + B_2) \le max(A_2 + B_2 + B_1, \ A_2 + A_1 + B_1) \\ \Leftrightarrow min(A_1, \ B_2) < min(A_2, \ B_1)$$

于是排序以后模拟一下就好了

 /**************************************************************
     Problem: 1727
     User: rausen
     Language: C++
     Result: Accepted
     Time:48 ms
     Memory:1156 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>
 using namespace std;
 const int N = 3e4 + ;
 const int inf = 1e9;

 inline int read();

 struct data {
     int a, b, k;

     inline void get() {
         a = read(), b = read();
         if (a < b) k = a;
         else k = inf - b;
     }

     inline bool operator < (const data &d) const {
         return k < d.k;
     }
 } a[N];

 int n, ans;

 int main() {
     int i, t1, t2;
     n = read();
     for (i = ; i <= n; ++i) a[i].get();
     sort(a + , a + n + );
     for (i = , t2 = ; i <= n; ++i) t2 += a[i].b;
     for (i = , t1 = ans = ; i <= n; ++i) {
         t1 += a[i].a;
         ans = max(ans, t1 + t2);
         t2 -= a[i].b;
     }
     printf("%d\n", ans);
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }