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 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? 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 cZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"pst">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 ?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 ?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
分析：
Pick定理：一个计算公式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。
如图：样例二
题目要求输出内部格点个数、边界格点个数、其面积。
1）.多边形面积求解公式：
2）. 多边形边界上的点数:
两顶点连线构成边界。
两顶点连线中（边界）所经过的点数即为，两顶点分别各自横纵坐标的差的最大公约数。
3）多边形内的点：
inside = area+1 - boundary/2;
#include
#include
#include
#include
#include
using namespace std; const int M = 1000 + 5; int area; int inside; int boundary; int x[M]; int y[M]; int p[M]; int q[M]; int boundary_work(int a, int b) { int t; while( b ) { t=b; b=a%b; a=t; } return a; } int area_work(int a, int b, int c, int d) { return (a*d-b*c); } int main() { int n, m; scanf("%d", &n); for(int cas=1; cas<=n; cas++) { scanf("%d", &m); p[0]=0; q[0]=0; inside=0; boundary=0; area=0; for(int i=1; i<=m; i++) { scanf("%d%d", &x[i], &y[i]); p[i]=p[i-1]+x[i]; q[i]=q[i-1]+y[i]; area += area_work( p[i-1], q[i-1], p[i], q[i] ); boundary += boundary_work( abs(x[i]), abs(y[i]) ); } inside = area/2+1-boundary/2; printf("Scenario #%d:\n%d %d %.1f\n\n", cas, inside, boundary, (double(area)/2)); } return 0; }