bzoj 1629: [Usaco2007 Demo]Cow Acrobats（贪心排序）

本文解析了Usaco2007Demo中的CowAcrobats问题，介绍了如何通过排序算法来确定最佳的牛堆叠顺序，以最小化堆叠过程中最高风险的值。通过实例演示了算法的具体实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1629: [Usaco2007 Demo]Cow Acrobats

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 1015 Solved: 528
[ Submit][ Status][ Discuss]

Description

Farmer John's N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away and join the circus. Their hoofed feet prevent them from tightrope walking and swinging from the trapeze (and their last attempt at firing a cow out of a cannon met with a dismal failure). Thus, they have decided to practice performing acrobatic stunts. The cows aren't terribly creative and have only come up with one acrobatic stunt: standing on top of each other to form a vertical stack of some height. The cows are trying to figure out the order in which they should arrange themselves within this stack. Each of the N cows has an associated weight (1 <= W_i <= 10,000) and strength (1 <= S_i <= 1,000,000,000). The risk of a cow collapsing is equal to the combined weight of all cows on top of her (not including her own weight, of course) minus her strength (so that a stronger cow has a lower risk). Your task is to determine an ordering of the cows that minimizes the greatest risk of collapse for any of the cows. //有三个头牛，下面三行二个数分别代表其体重及力量 //它们玩叠罗汉的游戏，每个牛的危险值等于它上面的牛的体重总和减去它的力量值，因为它要扛起上面所有的牛嘛. //求所有方案中危险值最大的最小

Input

* Line 1: A single line with the integer N. * Lines 2..N+1: Line i+1 describes cow i with two space-separated integers, W_i and S_i.

Output

* Line 1: A single integer, giving the largest risk of all the cows in any optimal ordering that minimizes the risk.

Sample Input

10 3

2 5

3 3

Sample Output

和http://blog.youkuaiyun.com/jaihk662/article/details/77824960一样

如果两头牛的重量和安全值分别为w1、y1；w2、y2

显然当y1-w2<y2-w1时第一头牛在第二头牛上面更优（危险值可以为负）

那么n条牛时按y1-w2<y2-w1的规则排序就好了

#include<stdio.h>
#include<algorithm>
using namespace std;
#define LL long long
typedef struct Res
{
	int w, y;
	bool operator < (const Res &b) const
	{
		if(y-b.w<b.y-w)
			return 1;
		return 0;
	}
}Res;
Res s[100005];
int main(void)
{
	LL ans, sum;
	int n, i;
	scanf("%d", &n);
	for(i=1;i<=n;i++)
		scanf("%d%d", &s[i].w, &s[i].y);
	sort(s+1, s+n+1);
	ans = -2147483647, sum = 0;
	for(i=1;i<=n;i++)
	{
		ans = max(ans, sum-s[i].y);
		sum += s[i].w;
	}
	printf("%lld\n", ans);
	return 0;
}