题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3007 Each person had do something foolish along with his or her growth.But,when he or she did this that time,they could not predict that this thing is a mistake and they will want this thing would rather…

Problem Description Each person had do something foolish along with his or her growth.But,when he or she did this that time,they could not predict that this thing is a mistake and they will want this thing would rather not happened.The world king Sco…

题意:给出n个点,求最小包围圆. 解法:这两天一直在学这个神奇的随机增量算法……看了这个http://soft.cs.tsinghua.edu.cn/blog/?q=node/1066之后自己写了好久一直写不对……后来在计算几何的模板上找到了…………orz膜拜一下 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h…

传送门 最小圆覆盖模板. //Achen #include<algorithm> #include<iostream> #include<cstring> #include<cstdlib> #include<vector> #include<cstdio> #include<queue> #include<cmath> ; typedef long long LL; typedef double db; us…

1. 题目描述有n个点,求能覆盖这n个点的半径最小的圆的圆心及半径. 2. 基本思路算法模板http://soft.cs.tsinghua.edu.cn/blog/?q=node/1066定义Di表示相对于P[1]和P[i]组成的最小覆盖圆,如果P[2..i-1]都在这个圆内,那么当前的圆心和半径即为最优解.如果P[j]不在这个圆内,那么P[j]一定在新的最小覆盖圆的边界上即P[1].P[j].P[i]组成的圆.因为三点可以确定一个圆,因此只需要不断的找到不满足的P[j],进而更新最优解即可.其…

Buried memory Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4067    Accepted Submission(s): 2171 Problem Description Each person had do something foolish along with his or her growth.But,when…

Hyperspace Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 1023    Accepted Submission(s): 492 Problem Description The great Mr.Smith has invented a hyperspace particle generator. The device i…

Hyperspace Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 314    Accepted Submission(s): 155 Problem Description The great Mr.Smith has invented a hyperspace particle generator. The device is…

有 n 种主武器,m 种副武器.每种武器有一个基础分数k种属性值 X[i] . 选出一种主武器 mw 和一种副武器 sw,使得两种武器的分数和 + 每个属性的差值尽量大.(参考下面的式子) 多维的最远曼哈顿距离. 因为对于每一种属性对答案的贡献,要么是 Xmw[i] - Xsw[i] ,要么是 Xsw[i] - Xmv[i]. 枚举每个主武器所有属性的符号,副武器的符号与主武器相反. 然后每次从主武器和副武器里面分别找一个属性值和基础分数的总和最大的,用这两个最大值的和更新答案. 因为k <=…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6153    Accepted Submission(s): 1457 Problem Description Memory units are numbered from 1 up to N. A sequence of memory units is called a memory…