侵略性的牛 题目大意:C头牛最大化他们的最短距离 常规题,二分法即可 #include <iostream> #include <algorithm> #include <functional> using namespace std; ]; bool judge(const int, const int,const int); void solve(const int, const int); int main(void) { int n, c; while (~sc…

原题如下: Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20524   Accepted: 9740 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions…

题目信息 作者:不详 链接:http://poj.org/problem?id=2456 来源:PKU JudgeOnline Aggressive cows[1] Time Limit: 1000MS Memory Limit: 65536K 描述 Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at p…

电话线 题目大意:一堆电话线要你接,现在有N个接口,总线已经在1端,要你想办法接到N端去,电话公司发好心免费送你几段不用拉网线,剩下的费用等于剩余最长电话线的长度,要你求出最小的费用. 这一看又是一个最小化最大值的问题(也可以看成是最大化最小值的问题),常规方法一样的就是把这个费用二分就好,但是这道题是道图论题,不一定经过所有的点,那我们就以二分基准长度为界限,把小于基准长度的那一部分看成是0,大于等于基准长度的看成是1,这样我们只用SPFA算法算最短路径就可以了,非常的巧妙 参考:http:/…

 挑选最美的珠宝 题目大意:挑选k个珠宝使得∑a/∑b最大,输出组合数 最大化平均值的标准题型,二分法就好了,一定要注意范围(10e-7),如果是10e-8就会tle,10e-6就是wa #include <iostream> #include <functional> #include <algorithm> using namespace std; struct _set { int v, w, num; }jewels[]; struct _out_set { d…

最大化平均值 题目大意:给定你n个分数,从中找出k个数,使∑a/∑b的最大值 这一题同样的也可以用二分法来做(用DP会超时,可见二分法是多么的实用呵!),大体上是这样子:假设最大的平均值是w,那么题目就是问存不存在∑a/b>=w,我们把这条式子变形 ∑a-w∑b>=0 那么这一题就变成了寻找k个最大的a-w*b,使∑a-w∑b>=0成立 #include <iostream> #include <algorithm> #include <functional…

缆绳大师 题目大意,把若干线段分成K份,求最大能分多长 二分法模型,C(x)就是题干的意思,在while那里做下文章就可以了,因为这个题目没有要求长度是整数,所以我们要不断二分才行,一般50-100次就可以了,精度足够了 最后要注意,这一题有四舍五入的陷阱,最好用floor处理一下 #include <iostream> #include <algorithm> #include <functional> #include <math.h> #define…

Monthly Expense 题目大意:不废话,最小化最大值 还是直接套模板,不过这次要注意,是最小化最大值,而不是最大化最小值,判断的时候要注意 联动3258 #include <iostream> #include <functional> #include <algorithm> using namespace std; ]; void Search(const int, const int, const int); bool C(const int, cons…

 去掉石头 题目大意:一群牛在河上的石头上跳来跳去,现在问你如何通过去掉M个石头,使得牛跳过石头的最短距离变得最大? 这一题比较经典,分治法的经典,二分法可以很方便处理这个问题,我们只要明白比较函数这个东西就可以了. 模板: while (……) { mid = (lb + rb) / ; if (Judge_C(……)) else rb = mid; } while判断条件可以根据是整形还是浮点型灵活变换,Judge_C就是比较函数,几乎所有的分治算法都可以这样归纳,我们只要找到合适的比较函数…

poj 2456 Aggressive cows && nyoj 疯牛 最大化最小值 二分 题目链接: nyoj : http://acm.nyist.net/JudgeOnline/problem.php?pid=586 poj : http://poj.org/problem?id=2456 思路: 二分答案,从前到后依次排放m头牛的位置,检查是否可行 代码: #include <iostream> #include <algorithm> #include &…

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7924   Accepted: 3959 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...…

题目传送门 /* 二分搜索:搜索安排最近牛的距离不小于d */ #include <cstdio> #include <algorithm> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; int x[MAXN]; int n, m; bool check(int d) { ; ; i<=m-; ++i) { ; while (cur <= n && x[…

题目传送门 POJ 2456 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000). His C (2 <= C <= N) cows don't like this…

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5436   Accepted: 2720 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...…

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11192   Accepted: 5492 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,..…

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20485   Accepted: 9719 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,..…

Aggressive cows Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22674 Accepted: 10636 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,x…

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18099   Accepted: 8619 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,..…

Aggressive cows 直接上中文了 Descriptions 农夫 John 建造了一座很长的畜栏,它包括N (2 <= N <= 100,000)个隔间,这些小隔间依次编号为x1,...,xN (0 <= xi <= 1,000,000,000). 但是,John的X (2 <= X <= N)头牛们并不喜欢这种布局,而且几头牛放在一个隔间里,他们就要发生争斗.为了不让牛互相伤害.John决定自己给牛分配隔间,使任意两头牛之间的最小距离尽可能的大,那么,这个…

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25944   Accepted: 11982 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,.…

原题链接:Aggressive cows 题目大意:农夫 建造了一座很长的畜栏,它包括  个隔间,这些小隔间依次编号为. 但是, 的  头牛们并不喜欢这种布局,而且几头牛放在一个隔间里,他们就要发生争斗.为了不让牛互相伤害. 决定自己给牛分配隔间,使任意两头牛之间的最小距离尽可能的大,那么,这个最大的最小距离是什么呢? 题目分析:题意想要表达的是 把头牛放到个带有编号的隔间里,使得任意两头牛所在的隔间编号的最小差值最大.这是一个典型的最小值最大化问题.先对畜栏编号从小到大排序,则最大距离不会超过…

///3.最大化最小值 /** POJ 2456 Aggressive cows Q:一排牛舍有N (2 <= N <= 100,000) 个,位置为x1,...,xN (0 <= xi <= 1,000,000,000) 为使牛之间不受到伤害,需最大化他们之间的距离,求最大化最近两头牛之间的距离 共M只牛 A: 条件C(x):可以使得最近两头牛之间的距离不小于d->求满足条件的最大d->如何高效的判断C(x) */ #include"iostream&quo…

POJ 2456 Agressive cows 农夫 John 建造了一座很长的畜栏,它包括N (2≤N≤100,000)个隔间,这 些小隔间的位置为x0,...,xN-1 (0≤xi≤1,000,000,000,均为整数,各不相同). John的C (2≤C≤N)头牛每头分到一个隔间.牛都希望互相离得远点省得 互相打扰.怎样才能使任意两头牛之间的最小距离尽可能的大,这个最 大的最小距离是多少呢 思想:二分,首先把输入的数据进行从小到大排序,再由最短距离为1,最长距离为(a[n-1] - a[0…

Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 23866   Accepted: 11141 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,.…

Aggressive cows http://poj.org/problem?id=2456 N间小屋,M头牛,使得牛跟牛之间的距离最远,以防止牛打架. 2<=N<=100000 2<=M<=N 0 <=xi<=109 ////////////////////////////////////////////////////////////// C(d):=可以安排牛的位置使得任意两头牛的间距都不小于d 使用二分搜索法解决: //参考文献:挑战程序设计大赛(第二版)/**…

Aggressive cows Time Limit: 1000MS Memory Limit: 65536K Total Submissions: Accepted: Description Farmer John has built a <= N <= ,) stalls. The stalls are located along a straight line at positions x1,...,xN ( <= xi <= ,,,). His C ( <= C &l…

POJ 2456 题意 农夫约翰有N间牛舍排在一条直线上,第i号牛舍在xi的位置,其中有C头牛对牛舍不满意,因此经常相互攻击.需要将这C头牛放在离其他牛尽可能远的牛舍,也就是求最大化最近两头牛之间的距离. 思路 二分搜索,现将牛舍排序,然后定义C(d),表示可安排的C头牛最近距离不小于d. #include <iostream> #include <algorithm> #include <cstdio> using namespace std; int N, C; /…

Aggressive cows 二分,关键是转化为二分! #include <cstdio> #include <algorithm> ; ; int N, C; int a[maxN]; using namespace std; bool judge(int x) { ; ; i < C;i++) { ; while (ctr < N && (a[ctr]-a[s] < x)) { ctr++; } if (ctr==N) { return fa…

最小最大...又是经典的二分答案做法.. -------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; ++i ) #def…

疯牛 时间限制:1000 ms  |  内存限制:65535 KB 难度:4   描述 农夫 John 建造了一座很长的畜栏,它包括N (2 <= N <= 100,000)个隔间,这些小隔间依次编号为x1,...,xN (0 <= xi <= 1,000,000,000).但是,John的C (2 <= C <= N)头牛们并不喜欢这种布局,而且几头牛放在一个隔间里,他们就要发生争斗.为了不让牛互相伤害.John决定自己给牛分配隔间,使任意两头牛之间的最小距离尽可能的…