Area

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

题目大意：

　　　　给一个平面上的简单多边形，求边上的点，多边形内的点，多边形面积。

解题思路：

　　　　Pick定理和叉积求多边形的面积。

　　　　Pick定理：一个计算点阵中顶点在格点上的多边形面积公式：S=a+b/2-1，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。

　　　  多边形面积：

1) △ABC的面积为向量AB与向量AC的叉乘的一半。

2)对于一个多边形，选定一个顶点P1，与其他顶点连线，可将多边形分为若干个三角形。

3)多边形面积为 abs(Sum{CrossMul(A,B,P1)|A,B为相邻的两个顶点}) (先求和再取abs，否则对于凹多边形会出错）

　　　 求在边上的顶点数：

　　　　　　对于Pa(x1,y1),Pb(x2,y2)所连成的选段，经过的格点的个数为Gcd(abs(x1-x2),abs(y1-y2))+1.

Code:

 1 #include<stdio.h> 2 #include<cmath> 3 #include<iostream> 4 #include<algorithm> 5 #define MAXN 10000 6 using namespace std; 7 int gcd(int a,int b) 8 { 9     return b==0?a:gcd(b,a%b);10 }11 int TeaTable(int x1,int y1,int x2,int y2)12 {13     return x1*y2-x2*y1;14 }15 int main()16 {17     int T,times=0;18     cin>>T;19     while (T--)20     {21         times++;22         int N;23         cin>>N;24         int area=0;25         int x=0,y=0,dx=0,dy=0;26         int cnt=0;27         for (int i=1;i<=N;i++)28         {29             int tx,ty;30             cin>>tx>>ty;31             dx=x+tx,dy=y+ty;32             cnt+=gcd(abs(tx),abs(ty));33             area+=TeaTable(x,y,dx,dy);34             x=dx,y=dy;35         }36         area=area>0?area:-area;37         double ans3=(double)area/2.0;38         int ans2=cnt;39         int ans1=(area+2-ans2)/2;40         printf("Scenario #%d:\n",times);41         printf("%d %d %.1lf\n\n",ans1,ans2,ans3);42     }43 44     return 0;45 }