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. 最后从遍历每一层的风险值,找到其中的最大值 我一开始对sum-p[i].a-p[i].b从小到大排序,这样第一次取出的就是能使最下层的牛的风险最小的方案,在上移一层时,这一层的风险值   为sum-p[i].a-p[i].b-p[0].a,由于p[0].a是固定值,所以第二次直接取出的就是能使该层的牛的风险最小的…

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 ca…

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 ca…

题目链接:click here~~ [题目大意] 给你n头牛叠罗汉.每头都有自己的重量w和力量s,承受的风险数rank就是该牛上面全部牛的总重量减去该牛自身的力量,题目要求设计一个方案使得全部牛里面风险最大的要最小. [解题思路]:依照w+s贪心放置,越大的(注意是w+s之和)越在以下.不难证明:假设最优放置时.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:rank1=sum-s1;第二头牛风险:rank2=sum+w1-…

一开始是往二分上去想的,如果risk是x,题目要求则可以转化为一个不等式,Si + x >= sigma Wj ,j表示安排在i号牛上面的牛的编号. 如果考虑最下面的牛那么就可以写成 Si + x  ≥ sum W - Wi,为了方便处理把i号牛的信息合并到一起 → Si + Wi + x ≥sum W. 二分x的时候,x是个常量,而从下面往上去安排牛的时候,下面的牛是没有影响决策的,可以看成把Wi去掉. 于是得到一个贪心的选法,把牛按照Si+Wi排序,从下面往上安排牛,可选择的牛应该满足Si+…

Silver Cow Party Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 26184 Accepted: 11963 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N…

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. The…

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\(x_1=sum-s_i\),$ x_2=sum+w_i-s_j$ 把\(j\)放在上面,危险值分别为\(x_3=sum-s_j\), \(x_4=sum+w_j-s_i\) 若把j放在上面更优,则有\(max(x_3,x_4)<max(x_1,x_2)\) 有四种情况 \(x_3<x_1\) \…

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 <= W_i <= 10,000)和力量(1 <= S_i <= 1,000,000,000).奶牛崩溃的风险等同于她身上所有的奶牛的重量(当然不包括她自己)减去她的力量.你的任务是确定奶牛的顺序,从而使得所有牛的风险中最大的一个尽量小.Input第1行:一个整数N 第2 . .N+ 1…

Cow Acrobats Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6253 Accepted: 2345 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 tightro…