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 research and development facility the company has installed the latest system of surveillance robots patrolling the area. These robots move along the walls of the facility and report suspicious observations to the central security office. The only flaw in the system a competitor抯 agent could find is the fact that the robots radio their movements unencrypted. Not being able to find out more, the agent wants to use that information to calculate the exact size of the area occupied by the new facility. It is public knowledge that all the corners of the building are situated on a rectangular grid and that only straight walls are used. Figure 1 shows the course of a robot around an example area.



Figure 1: Example area.

You are hired to write a program that calculates the area occupied
by the new facility from the movements of a robot along its walls. You
can assume that this area is a polygon with corners on a rectangular
grid. However, your boss insists that you use a formula he is so proud
to have found somewhere. The formula relates the number I of grid points
inside the polygon, the number E of grid points on the edges, and the
total area A of the polygon. Unfortunately, you have lost the sheet on
which he had written down that simple formula for you, so your first
task is to find the formula yourself.

Input

The first line contains the number of scenarios.

For each scenario, you are given the number m, 3 <= m < 100,
of movements of the robot in the first line. The following m lines
contain pairs 揹x dy�of integers, separated by a single blank, satisfying
.-100 <= dx, dy <= 100 and (dx, dy) != (0, 0). Such a pair means
that the robot moves on to a grid point dx units to the right and dy
units upwards on the grid (with respect to the current position). You
can assume that the curve along which the robot moves is closed and that
it does not intersect or even touch itself except for the start and end
points. The robot moves anti-clockwise around the building, so the area
to be calculated lies to the left of the curve. It is known in advance
that the whole polygon would fit into a square on the grid with a side
length of 100 units.

Output

The
output for every scenario begins with a line containing 揝cenario #i:�
where i is the number of the scenario starting at 1. Then print a single
line containing I, E, and A, the area A rounded to one digit after the
decimal point. Separate the three numbers by two single blanks.
Terminate the output for the scenario with a blank line.

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

Source

 

【思路】

Pick定理

同上题。

【代码】

 #include<cstdio>
using namespace std; struct Pt {
int x,y;
Pt(int x=,int y=) :x(x),y(y){};
};
Pt p[];
typedef Pt vec;
vec operator - (Pt a,Pt b) { return vec(a.x-b.x,a.y-b.y); } int abs(int x) { return x<? -x:x; }
int gcd(int a,int b) { return b==? a:gcd(b,a%b); }
int cross(Pt a,Pt b) { return a.x*b.y-a.y*b.x; }
int calc(Pt a,Pt b) { return gcd(abs(a.x-b.x),abs(a.y-b.y)); } int n;
int main() {
int T,kase=;
scanf("%d",&T);
while(T--) {
scanf("%d",&n);
n++;
p[]=Pt(,);
for(int i=;i<n;i++) {
scanf("%d%d",&p[i].x,&p[i].y);
p[i].x+=p[i-].x , p[i].y+=p[i-].y;
}
int S=,b=;
for(int i=;i<n-;i++) {
S += cross(p[i],p[i+]);
b += calc(p[i],p[i+]);
}
b += calc(p[n-],p[])+calc(p[],p[]);
printf("Scenario #%d:\n%d %d %.1f\n\n",++kase,(S-b+)/,b,0.5*S);
}
return ;
}

poj 1265 Area(Pick定理)的更多相关文章

  1. poj 1265 Area (Pick定理+求面积)

    链接:http://poj.org/problem?id=1265 Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions:  ...

  2. POJ 1265 Area (Pick定理 & 多边形面积)

    题目链接:POJ 1265 Problem Description Being well known for its highly innovative products, Merck would d ...

  3. poj 1265 Area(pick定理)

    Area Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4373 Accepted: 1983 Description Bein ...

  4. [poj 1265]Area[Pick定理][三角剖分]

    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 ...

  5. poj 1265 Area( pick 定理 )

    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 ...

  6. Area - POJ 1265(pick定理求格点数+求多边形面积)

    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 ...

  7. poj 1265 Area 面积+多边形内点数

    Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5861   Accepted: 2612 Description ...

  8. POJ 1265 Area POJ 2954 Triangle Pick定理

    Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5227   Accepted: 2342 Description ...

  9. POJ 1265 Area (pick定理)

    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 ...

随机推荐

  1. (转)IOS中获取各种文件的目录路径的方法

    iphone沙箱模型的有四个文件夹,分别是什么,永久数据存储一般放在什么位置,得到模拟器的路径的简单方式是什么. documents,tmp,app,Library. (NSHomeDirectory ...

  2. Java中“|”和“||”用法的区别

    例子: int a = 5; int b = 10; if(a > 4 | b++ > 10) { System.out.println("a:"+a+"\n ...

  3. #添加屏蔽IP LINUX

    netfilter/iptables 的最大优点是它可以配置有状态的防火墙,这是 ipfwadm 和 ipchains 等以前的工具都无法提供的一种重要功能.有状态的防火墙能够指定并记住为发送或接收信 ...

  4. 【实习记】2014-09-24万事达卡bin查询项目总结

            8月28号,接到这个问题:现有前缀查询速度较慢,改进此知值求区间问题. 一开始没想到用二分法,更没有想到这个项目用了一个月,这一个月里,我学习并使用了middle框架写出了server ...

  5. 【DP_树形DP专题】题单总结

    转载自 http://blog.csdn.net/woshi250hua/article/details/7644959#t2 题单:http://vjudge.net/contest/123963# ...

  6. Razor与ASPX语法比较

  7. Using .NET 4's Lazy<T> 实现单实例

    Using .NET 4's Lazy<T> type Explanation of the following code: If you're using .NET 4 (or high ...

  8. css常用伪类记录

    1.超链接使用css伪类设置颜色 a:link {color: #000000} /* 未访问的链接 */a:visited {color: #d90a81} /* 已访问的链接 */a:hover ...

  9. leetcode其余题目

    1.Largest Rectangle in Histogram http://discuss.leetcode.com/questions/259/largest-rectangle-in-hist ...

  10. python 重载 __hash__ __eq__

    __author__ = 'root' from urlparse import urlparse class host_news(): def __init__(self, id, url): se ...