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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
1 /************************************************************** 2 Problem: 1727 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:48 ms 7 Memory:1156 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 using namespace std; 13 const int N = 3e4 + 5; 14 const int inf = 1e9; 15 16 inline int read(); 17 18 struct data { 19 int a, b, k; 20 21 inline void get() { 22 a = read(), b = read(); 23 if (a < b) k = a; 24 else k = inf - b; 25 } 26 27 inline bool operator < (const data &d) const { 28 return k < d.k; 29 } 30 } a[N]; 31 32 int n, ans; 33 34 int main() { 35 int i, t1, t2; 36 n = read(); 37 for (i = 1; i <= n; ++i) a[i].get(); 38 sort(a + 1, a + n + 1); 39 for (i = 1, t2 = 0; i <= n; ++i) t2 += a[i].b; 40 for (i = 1, t1 = ans = 0; i <= n; ++i) { 41 t1 += a[i].a; 42 ans = max(ans, t1 + t2); 43 t2 -= a[i].b; 44 } 45 printf("%d\n", ans); 46 return 0; 47 } 48 49 inline int read() { 50 static int x; 51 static char ch; 52 x = 0, ch = getchar(); 53 while (ch < ‘0‘ || ‘9‘ < ch) 54 ch = getchar(); 55 while (‘0‘ <= ch && ch <= ‘9‘) { 56 x = x * 10 + ch - ‘0‘; 57 ch = getchar(); 58 } 59 return x; 60 }
BZOJ1727 [Usaco2006 Open]The Milk Queue 挤奶队列
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原文地址:http://www.cnblogs.com/rausen/p/4461589.html