链接:http://poj.org/problem?id=1265 Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4969   Accepted: 2231 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionag…

题目链接:POJ 1265 Problem Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research and development facility the company has installed the latest s…

Area Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4373 Accepted: 1983 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research a…

题意: 给出机器人移动的向量, 计算包围区域的内部整点, 边上整点, 面积. 思路: 面积是用三角剖分, 边上整点与GCD有关, 内部整点套用Pick定理. S = I + E / 2 - 1 I 为内整点数, E为边界整点数, S为面积. Separate the three numbers by two single blanks.....好吧, 理解成中间空两格PE一次> < #include <cstdio> #include <cstring> #includ…

题目:http://poj.org/problem?id=1265 题意:已知机器人行走步数及每一步的坐标   变化量 ,求机器人所走路径围成的多边形的面积.多边形边上和内部的点的数量. 思路:1.以格子点为顶点的线段,覆盖的点的个数为GCD(dx,dy),其中,dxdy分别为线段横向占的点数和纵向占的点数.如果dx或dy为0,则覆盖的点数为dy或dx. 2.Pick公式:平面上以格子点为顶点的简单多边形,如果边上的点数为on,内部的点数为in,则它的面积为A=on/2+in-1. 3.任意一个…

题目大意:以原点为起点然后每次增加一个x,y的值,求出来最后在多边形边上的点有多少个,内部的点有多少个,多边形的面积是多少. 分析: 1.以格子点为顶点的线段,覆盖的点的个数为GCD(dx,dy),其中,dxdy分别为线段横向占的点数和纵向占的点数.如果dx或dy为0,则覆盖的点数为dy或dx.2.Pick公式:平面上以格子点为顶点的简单多边形的面积=边上的点数/2+内部的点数+1.3.任意一个多边形的面积等于按顺序求相邻两个点与原点组成的向量的叉积之和. 代码如下: -------------…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5861   Accepted: 2612 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5227   Accepted: 2342 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5666   Accepted: 2533 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

题目大意:已知机器人行走步数及每一步的坐标变化量,求机器人所走路径围成的多边形的面积.多边形边上和内部的点的数量. 思路:叉积求面积,pick定理求点. pick定理:面积=内部点数+边上点数/2-1. // Time 0ms; Memory 236K #include<iostream> #include<cstdio> #include<cmath> using namespace std; struct point { int x,y; point(int xx=…

http://poj.org/problem?id=1265 题意:起始为(0,0),给出每个点的偏移量,求依次连接这些点形成的多边形边界上格点的个数. 思路:先将各个点的坐标求出存入,由pick定理知: 面积 = 内部格点数目+边上格点数目/2-1: 每条边边格点数目 = gcd(x2-x1,y2-y1): 内部格点数目 = 面积+1-边界格点数目/2: #include <stdio.h> #include <stdlib.h> #include <math.h>…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5429   Accepted: 2436 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

Area DescriptionBeing well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research and development facility the company has installed the latest system of surveilla…

题目:http://poj.org/problem?id=1265 Sample Input 2 4 1 0 0 1 -1 0 0 -1 7 5 0 1 3 -2 2 -1 0 0 -3 -3 1 0 -3 Sample Output Scenario #1: 0 4 1.0 Scenario #2: 12 16 19.0 注意:题目给出的成对的数可不是坐标,是在x和y方向走的数量. 边界上的格点数:一条左开右闭的线段(x1, x2)->(x2, y2)上的格点数为:gcd( abs(x2-x1…

有一种定理,叫毕克定理....                             Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4352   Accepted: 1977 Description Being well known for its highly innovative products, Merck would definitely be a good target for industria…

Area Time Limit: 1000MS Memory Limit: 10000K Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research and development facility the company has…

题目大意:给出三个点的坐标,问在这三个点坐标里面的整数坐标点有多少个(不包含边上的) 匹克定理:I = (A-E) / 2 + 1; A: 表示多边形面积 I : 表示多边形内部的点的个数 E: 表示在多边形上的点的个数 // Time 0ms; Memory 164K #include<iostream> #include<cstdio> #include<cmath> using namespace std; typedef struct point { int x…

PICK定理: S=I+O/2-1 S为多边形面积,I多边形内部的格点,O是多边形边上的格点 其中边上格点求法: 假设两个点A(x1,y1),B(x2,y2) 线段AB间格点个数为gcd(abs(x1-x2),abs(y1-y2))-1 特判x1-x2==0 或者 y1-y2==0,则覆盖的点数为 y2-y1 或 x2-x1 POJ 2594 Triangle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5106  …

Area POJ - 1265 皮克定理是指一个计算点阵中顶点在格点上的多边形面积公式,该公式可以表示为2S=2a+b-2, 其中a表示多边形内部的点数,b表示多边形边界上的点数,S表示多边形的面积. 适用范围:必须是格点多边形.S = A / 2 + B - 1 #include<stdio.h> #include<cmath> #include<cstring> #include<algorithm> using namespace std; #defi…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5811   Accepted: 2589 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

http://poj.org/problem?id=2954 表示我交了20+次... 为什么呢?因为多组数据我是这样判断的:da=sum{a[i].x+a[i].y},然后!da就表示没有数据了QAQ我居然查了如此久都没查出来!!!!注意负数啊负数啊啊啊啊啊啊啊 本题是pick定理:当多边形的顶点均为整数时,面积=内部整点+边上整点/2-1 然后本题要求内部整点 #include <cstdio> #include <cstring> #include <cmath>…

链接:http://poj.org/problem?id=2954 Triangle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5043   Accepted: 2164 Description A lattice point is an ordered pair (x, y) where x and y are both integers. Given the coordinates of the vertices…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4725   Accepted: 2135 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

什么是Pick定理(皮克定理) 来自wiki的介绍: 给定顶点座标均是整点(或正方形格子点)的简单多边形,皮克定理说明了其面积 \(A\)和内部格点数目 \(i\) .边上格点数目 \(b\) 的关系:\(A = i + \frac b 2 - 1\). 因为所有简单多边形都可切割为一个三角形和另一个简单多边形.考虑一个简单多边形 \(P\),及跟\(P\)有一条共同边的三角形\(T\).若\(P\) 符合皮克公式,则只要证明\(P\)加上\(T\) 的\(PT\)亦符合皮克公式(I),与及三角…

//pick定理:面积=内部整数点数+边上整数点数/2-1 // POJ 2954 #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h> using namespace std; #define LL long long typedef pair<int,i…

链接  Pick定理是说,在一个平面直角坐标系内,如果一个多边形的顶点全都在格点上,那么这个图形的面积恰好就等于边界上经过的格点数的一半加上内部所含格点数再减一. pick定理的一些应用 题意不好懂,给出的x,y并不是坐标而是向x轴方向y轴方向移动的距离. #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #i…

这种1A的感觉真好 #include <cstdio> #include <vector> #include <cmath> using namespace std; typedef long long LL; struct Point { LL x, y; Point(LL x=, LL y=):x(x), y(y) {} }; Point operator - (const Point& A, const Point& B) { return Poi…

P1150 - 绳子围点 From 332404521    Normal (OI)总时限:10s    内存限制:128MB    代码长度限制:64KB 背景 Background 最近小小鱼在研究平面几何,遇到一个难题,怎么也想不出来,于是找到大牛你来帮他做. 描述 Description 给出平面上n个点,所有点的坐标都是整数,小小鱼用一条绳子围成一个封闭图形,把这些点全部围在里面,并且所用绳子长度最短.围好了之后,小小鱼想知道这条绳子总共围住了多少个横.纵坐标都为整数的点(包括给出的n…

一.概念 假设P的内部有I(P)个格点,边界上有B(P)个格点,则P的面积A(P)为:A(P)=I(P)+B(P)/2-1. 二.说明 Pick定理主要是计算格点多边形(定点全是格点的不自交图形)P的面积与其边界和内部格点数之间的关系. 格点多边形的面积A(P)可以通过叉积计算出来,不过叉积计算出来的面积是实际面积的2倍: 边界上的格点B(P)可以通过计算相邻两点的横坐标之差与纵坐标之差的最大公约数的和得到: 内部的格点I(P)则通过公式得:I(P) = A(P)-B(P)/2+1计算出. 解释…

一.概念 假设P的内部有I(P)个格点,边界上有B(P)个格点,则P的面积A(P)为:A(P)=I(P)+B(P)/2-1. 二.说明 Pick定理主要是计算格点多边形(定点全是格点的不自交图形)P的面积与其边界和内部格点数之间的关系. 格点多边形的面积A(P)可以通过叉积计算出来,不过叉积计算出来的面积是实际面积的2倍: 边界上的格点B(P)可以通过计算相邻两点的横坐标之差与纵坐标之差的最大公约数的和得到: 内部的格点I(P)则通过公式得:I(P) = A(P)-B(P)/2+1计算出. 解释…