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=-INF. #include<iostream> #include<cstdio> #include<queue> #include<cstring> #include<algorithm> #include<cmath> #includ…

描述 http://poj.org/problem?id=2456 有n个小屋,线性排列在不同位置,m头牛,每头牛占据一个小屋,求最近的两头牛之间距离的最大值. Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10095   Accepted: 4997 Description Farmer John has built a new long barn, with N (2 <= N <=…

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 \(x_3<x_1\) \…

River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9923   Accepted: 4252 Description Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. T…

Cow Acrobats Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4998   Accepted: 1892 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 tig…

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…

描述 http://poj.org/problem?id=3258 给出起点和终点之间的距离L,中间有n个石子,给出第i个石子与起点之间的距离d[i],现在要去掉m个石子(不包括起终点),求距离最近的两个石子(包括起终点)之间距离的最大值. River Hopscotch Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10841   Accepted: 4654 Description Every year the co…

题目链接:http://poj.org/problem?id=3258 题意:给n个石头,起点和终点也是两个石头,去掉这石头中的m个,使得石头间距的最小值最大. 思路:二分石头间的最短距离,每次贪心地check一下是否满足条件即可,具体看代码. AC代码: #include<iostream> #include<stack> #include<vector> #include<algorithm> #include<cmath> using na…

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险值 分析:这就是noip2012 T2的来源= =只不过这里是加,noip里是乘 不妨设所有牛都按最优顺序排好了,考虑相邻的两头牛i和i+1,如果交换他们的位置,那么对前面和后面的结果都无影响,只是他们两个的风险值变化了(变大了),于是我们可以得到这个时候i和i+1的关系 设w1+w2+...+wi-1…

题目并不难,就是比赛的时候没敢去二分,也算是一个告诫,应该敢于思考…… #include<stdio.h> #include<iostream> using namespace std; int main() { long long n; scanf("%I64d",&n); ,right=1e18,mid,num,m,s; ; while(left<=right) { mid=(left+right)>>; num = ; ;i &l…