POJ 3045 Cow Acrobats （贪心）

最新推荐文章于 2021-03-03 22:12:20 发布

原创最新推荐文章于 2021-03-03 22:12:20 发布 · 557 阅读

1 ·

CC 4.0 BY-SA版权

挑战程序设计竞赛(第2版) 习题例题解同时被 2 个专栏收录

76 篇文章

订阅专栏

搜索 - 二/三分搜索

26 篇文章

订阅专栏

本文解析了CowAcrobats问题，介绍了如何通过贪心算法确定牛堆叠顺序以最小化倒塌风险。通过将牛按体重加力量排序，并计算每头牛的风险值，最终找到风险最小化的排列方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Cow Acrobats

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3192		Accepted: 1260

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 ithin 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.

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

Sample Output

这题 - - 没怎么懂转载一发题解贪心真是弱啊

按照w+s贪心叠，越大的越在下面。如果最优放置时，相邻两头牛属性分别为w1,s1,w2,s2,第一头牛在第二头上面，sum为第一头牛上面的牛的体重之和，那么

第一头牛风险：a=sum-s1;第二头牛风险:b=sum+w1-s2;

交换两头牛位置之后 a'=sum+w2-s1,b'=sum-s2，

由于是最优放置，所以w2-s1>=w1-s2，即w2+s2>=w1+s1，所以和最大的就该老实的在下面呆着= =！

AC代码如下：

//
//  POJ 3045 Cow Acrobats
//
//  Created by TaoSama on 2015-04-25
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n;
struct Cow {
    int w, s;
    bool operator < (const Cow& rhs) const {
        return w + s < rhs.w + rhs.s;
    }
} a[50005];

int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%d%d", &a[i].w, &a[i].s);
    sort(a + 1, a + 1 + n);
    int ans = -INF, sum = 0;
    for(int i = 1; i <= n; ++i){
		ans = max(ans, sum - a[i].s);
		sum += a[i].w;
    }
    printf("%d\n", ans);
    return 0;
}