Lifting the Stone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6104    Accepted Submission(s): 2546 Problem Description There are many secret openings in the floor which are covered by a big…

题目链接:http://poj.org/problem?id=1385 题目大意:给你一个多边形的点,求重心. 首先,三角形的重心: ( (x1+x2+x3)/3 , (y1+y2+y3)/3 ) 然后多边形的重心就是将多边形划分成很多个三角形,以三角形面积为权值,将每个三角形的重心加权平均. 注意: pair<double,double>会MLE.. fabs会损失精度?(这个我也不知道),因此在用向量叉积求三角形面积的时候最好是直接让面积求出来就是正的..否则fabs就WA了... 代码:…

Lifting the Stone Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 230 Accepted Submission(s): 130   Problem Description There are many secret openings in the floor which are covered by a big heavy…

Lifting the Stone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4819    Accepted Submission(s): 2006 Problem Description There are many secret openings in the floor which are covered by a big…

转载请注明出处:http://blog.csdn.net/u012860063 题目链接:pid=1115">http://acm.hdu.edu.cn/showproblem.php? pid=1115 Lifting the Stone Problem Description There are many secret openings in the floor which are covered by a big heavy stone. When the stone is lift…

Lifting the Stone 题目链接: http://acm.hust.edu.cn/vjudge/contest/130510#problem/G Description There are many secret openings in the floor which are covered by a big heavy stone. When the stone is lifted up, a special mechanism detects this and activates…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4764 题目大意:Tang和Jiang玩石子游戏,给定n个石子,每次取[1,k]个石子,最先取完的人失败,Tang先取,求博弈. Sample Input 1 1 30 3 10 2 0 0   Sample Output Jiang Tang Jiang 分析:可以看成是谁先取完n-1个石子,谁获胜,则变成完完全全的巴什博弈,当然要考虑特殊情况. 代码如下: #include<iostream>…

链接:https://ac.nowcoder.com/acm/contest/893/D来源:牛客网 题目描述 有n堆石子排成一排,第i堆石子有aiai个石子. 每次,你可以选择任意相邻的两堆石子进行合并,合并后的石子数量为两堆石子的和,消耗的体力等价于两堆石子中石子数少的那个. 请问,将所有的石子合并成一堆,你所消耗的体力最小是多少? 输入描述: 第一行是一个整数T(1≤T≤20)T(1≤T≤20),表示样例的个数.每个样例的第一行是一个整数n(1≤n≤10000)n(1≤n≤10000),表…

题意:有n个石子堆,每一个都可以轮流做如下操作:选一个石堆,移除至少1个石子,然后可以把这堆石子随便拿几次,随便放到任意的其他石子数不为0的石子堆,也可以不拿.不能操作败. 思路:我们先来证明,如果某个石子数有偶数堆,则先手必败,因为无论先手怎么做,后手都能模仿先手,最后把石子取光.显然全是偶数堆是必败态.如果有奇数堆怎么办?我们就把最大的奇数堆取光,然后把其他奇数堆变成偶数堆.但是一定能保证可以吗?答案是可以.假设奇数堆的石子数为 x1,x2,x3...xn,那么我们分别给每一堆加上x2-x1…

[BZOJ2138]stone(线段树,Hall定理) 题面 BZOJ 题解 考虑一个暴力. 我们对于每堆石子和每个询问,显然是匹配的操作. 所以可以把石子拆成\(a_i\)个,询问点拆成\(K_i\)个,这样就是每次进行一次二分图的匹配. 当然可以用网络流+线段树优化连边来做,但是这样复杂度太高. 还是回到二分图的匹配问题,我们现在要验证的就是是否存在对于当前询问点的完美匹配. 关于完美匹配,有\(Hall\)定理,如果存在完美匹配,假设左侧的点有\(|X|\)个,那么这些点连向右边的点的点集…