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 + 5;
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 = 1; i <= n; ++i) a[i].get();
    sort(a + 1, a + n + 1);
    for (i = 1, t2 = 0; i <= n; ++i) t2 += a[i].b;
    for (i = 1, t1 = ans = 0; i <= n; ++i) {
        t1 += a[i].a;
        ans = max(ans, t1 + t2);
        t2 -= a[i].b;
    }
    printf("%d
", ans);
    return 0;
}
 
inline int read() {
    static int x;
    static char ch;
    x = 0, ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}