Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11539   Accepted: 3395   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory…

Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7632   Accepted: 2263   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory…

Bridge Across Islands Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10455   Accepted: 3093   Special Judge Description Thousands of thousands years ago there was a small kingdom located in the middle of the Pacific Ocean. The territory…

链接: http://poj.org/problem?id=2187 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/E Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24254   Accepted: 7403 Description Bessie, Farmer John's prize cow, h…

给定点集的最远两点的距离. 先用graham求凸包.旋(xuán)转(zhuàn)卡(qiǎ)壳(ké)求凸包直径. ps:旋转卡壳算法的典型运用 http://blog.csdn.net/hanchengxi/article/details/8639476. #include <cstdio> #include <cmath> #include <algorithm> #define sqr(x) (x)*(x) #define N 50001 using names…

1069: [SCOI2007]最大土地面积 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 3629  Solved: 1432[Submit][Status][Discuss] Description 在某块平面土地上有N个点,你可以选择其中的任意四个点,将这片土地围起来,当然,你希望这四个点围成的多边形面积最大. Input 第1行一个正整数N,接下来N行,每行2个数x,y,表示该点的横坐标和纵坐标. Output 最大的多边形面积,答案精确到…

一些基本的定义在这里: [模板] 计算几何1(基础): 点/向量/线/圆/多边形/其他运算 自适应Simpson Simpson's Rule: \[ \int ^b_a f(x)dx\approx \frac{b-a}6(f(a)+4f(\frac{a+b}{2})+f(b)) \] 这是对二次函数的积分估值, 对于一, 二次函数来说都是准确的. 但是对于其他函数来说, 这只是利用二次函数进行近似. 可以采用自适应精度的手段, 使得估值接近真实结果. 详见代码. 然后这是误差估计, 详见 ad…

Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points. Input The input consists of several test cases. The first line of each test case contains an integer n, indica…

给两个凸包,求这两个凸包间最短距离 旋转卡壳的基础题 因为是初学旋转卡壳,所以找了别人的代码进行观摩..然而发现很有意思的现象 比如说这个代码(只截取了关键部分) double solve(Point* P, Point* Q, int n, int m) { , ymaxQ = ; ; i < n; ++i) if (P[i].y < P[yminP].y) yminP = i; // P上y坐标最小的顶点 ; i < m; ++i) if (Q[i].y > Q[ymaxQ].…

题目链接:http://poj.org/problem?id=2187 旋转卡壳算法:http://www.cppblog.com/staryjy/archive/2009/11/19/101412.html 或 http://cgm.cs.mcgill.ca/~orm/rotcal.frame.html #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #includ…