题目链接:http://poj.org/problem?id=3045

Cow Acrobats
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 5713   Accepted: 2151

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

3
10 3
2 5
3 3

Sample Output

2

Hint

OUTPUT DETAILS:

Put the cow with weight 10 on the bottom. She will carry the other two cows, so the risk of her collapsing is 2+3-3=2. The other cows have lower risk of collapsing.

Source

 
 
 
题解:
2.自己的思考:根据承受力来排序,体重可能会走向极端;根据体重来排序,承受力也可能会走向极端。所以片面的考虑是得不到结果的(做题都能映射出人生,还能说些什么),既然体重和承受力共同影响这结果,所以就需要综合两者来考虑,即两者之和。
 
 
代码如下:
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e5+; struct node
{
int w, s;
bool operator<(const node &x)const{
return (w+s)<(x.w+x.s);
}
}a[MAXN]; int main()
{
int n;
while(scanf("%d", &n)!=EOF)
{
for(int i = ; i<=n; i++)
scanf("%d%d", &a[i].w, &a[i].s);
sort(a+, a++n);
LL ans = -INF, tot = ;
for(int i = ; i<=n; i++)
{
ans = max(ans, tot-a[i].s);
tot += a[i].w;
}
printf("%lld\n", ans);
}
}

POJ3045 Cow Acrobats —— 思维证明的更多相关文章

  1. poj3045 Cow Acrobats (思维,贪心)

    题目: poj3045 Cow Acrobats 解析: 贪心题,类似于国王游戏 考虑两个相邻的牛\(i\),\(j\) 设他们上面的牛的重量一共为\(sum\) 把\(i\)放在上面,危险值分别为\ ...

  2. POJ3045 Cow Acrobats 2017-05-11 18:06 31人阅读 评论(0) 收藏

    Cow Acrobats Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4998   Accepted: 1892 Desc ...

  3. POJ-3045 Cow Acrobats (C++ 贪心)

    Description Farmer John's N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away a ...

  4. [USACO2005][POJ3045]Cow Acrobats(贪心)

    题目:http://poj.org/problem?id=3045 题意:每个牛都有一个wi和si,试将他们排序,每头牛的风险值等于前面所有牛的wj(j<i)之和-si,求风险值最大的牛的最小风 ...

  5. poj3045 Cow Acrobats(二分最大化最小值)

    https://vjudge.net/problem/POJ-3045 读题后提取到一点:例如对最底层的牛来说,它的崩溃风险=所有牛的重量-(底层牛的w+s),则w+s越大,越在底层. 注意范围lb= ...

  6. POJ3045 Cow Acrobats

    题意 Farmer John's N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away and join t ...

  7. 【POJ - 3045】Cow Acrobats (贪心)

    Cow Acrobats Descriptions 农夫的N只牛(1<=n<=50,000)决定练习特技表演. 特技表演如下:站在对方的头顶上,形成一个垂直的高度. 每头牛都有重量(1 & ...

  8. BZOJ1629: [Usaco2007 Demo]Cow Acrobats

    1629: [Usaco2007 Demo]Cow Acrobats Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 601  Solved: 305[Su ...

  9. POJ 3045 Cow Acrobats (贪心)

    POJ 3045 Cow Acrobats 这是个贪心的题目,和网上的很多题解略有不同,我的贪心是从最下层开始,每次找到能使该层的牛的风险最小的方案, 记录风险值,上移一层,继续贪心. 最后从遍历每一 ...

随机推荐

  1. spring mvc 单元测试示例

    import java.awt.print.Printable; import java.io.IOException; import javax.servlet.http.HttpServletRe ...

  2. uva 11916 解模方程a^x=b (mod n)

      Emoogle Grid  You have to color an M x N ( 1M, N108) two dimensional grid. You will be provided K  ...

  3. substring用法

    字符串截取,substring(int beginIndex) 返回一个新的字符串,它是此字符串的一个子字符串. substring(int beginIndex, int endIndex) 返回一 ...

  4. bzoj 2694: Lcm

    2694: Lcm Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 422  Solved: 220[Submit][Status][Discuss] ...

  5. Java中获取项目根路径和类加载路径的7种方法

    引言 在web项目开发过程中,可能会经常遇到要获取项目根路径的情况,那接下来我就总结一下,java中获取项目根路径的7种方法,主要是通过thisClass和System,线程和request等方法. ...

  6. 基于SSH+shiro+solr的家庭记账系统

    项目地址: https://github.com/jianghuxiaoao/homeaccount

  7. IntelliJ IDEA出现:This file is indented with tabs instead of 4 spaces的问题解决

    根据阿里巴巴Java开发手册,不能使用Tab字符,改成4个字符,设置如下: 注意:是不选择! 一定要选择这个:

  8. CSS 居中 可随着浏览器变大变小而居中

    关键代码: 外部DIV使用: text-align:center; 内部DIV使用: margin-left:auto;margin-right:auto 例: <div style=" ...

  9. ganglia-monitoring-centos-linux

    http://www.tecmint.com/install-configure-ganglia-monitoring-centos-linux/ http://monitor.gitlab.net/ ...

  10. mysqldump 把数据库备份到异地的服务器

    原文:http://www.open-open.com/code/view/1420121471484 这个方法可以把通过mysqldump 把本地数据库备份到远端主机, 中间数据的传输通过 ssh ...