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: -------------…

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…

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…

链接:http://poj.org/problem?id=1654 Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14952   Accepted: 4189 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orth…

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=…

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…

//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…

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>…

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…

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…

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 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…

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…

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…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17456   Accepted: 4847 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

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…

题目大意: 默认从零点开始 给定n次x y上的移动距离 组成一个n边形(可能为凹多边形) 输出其 内部整点数 边上整点数 面积 皮克定理 多边形面积s = 其内部整点in + 其边上整点li / 2 - 1 那么求内部整点就是 in = s + 1 - li / 2 求整个多边形边上的整点数 //求两点ab之间的整点数 int lineSeg(P a,P b) { int dx=abs(a.x-b.x), dy=abs(a.y-b.y); && dy==) ; ; // 不包括a b两个顶…

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…

链接:http://poj.org/problem?id=3348 Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6677   Accepted: 3020 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with…

/* poj 1654 Area 多边形面积 题目意思很简单,但是1000000的point开不了 */ #include<stdio.h> #include<math.h> #include<string.h> const int N=1000000+10; const double eps=1e-8; struct point { double x,y; point(){} point(double a,double b):x(a),y(b){} }; int le…

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…

LINK 题意:给出点集,求凸包的面积 思路:主要是求面积的考察,固定一个点顺序枚举两个点叉积求三角形面积和除2即可 /** @Date : 2017-07-19 16:07:11 * @FileName: POJ 3348 凸包面积 叉积.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <…