Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 6319 | Accepted: 2771 |
Description
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
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
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
题意:给出一个点阵,上面的点连成了一个多边形,求多边形内部的点,多边形边界上的点,还有多边形的面积
注意,给的数是相对上一个点移动距离,也就是这一个点和上一个点的向量
分析:1.pick定理,S=a/2+b-1 S是多边形的面积,a是多边形边界上的点的数量,b是多边形内点的数量
2.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
3.多边形边界上点个数等于gcd(x,y) x是相邻两个点横座标之差的绝对值,y是相邻两点纵座标之差的绝对值
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
using namespace std;
struct node
{
int x,y;
}q[150];
int gcd(int a,int b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
double area(node a,node b)
{
return a.x*b.y-a.y*b.x;
}
int main()
{
int n,m,i,j,t;
int xx,yy;
scanf("%d",&t);
for(int ll=1;ll<=t;ll++)
{
int bd=0;
double s=0;
scanf("%d",&n);
q[0].x=0;
q[0].y=0;
for(i=1;i<=n;i++)
{
scanf("%d%d",&xx,&yy);
bd+=gcd(abs(xx),abs(yy));
q[i].x=q[i-1].x+xx;
q[i].y=q[i-1].y+yy;
s+=area(q[i],q[i-1]);
}
s=fabs(s);
printf("Scenario #%d:\n",ll);
printf("%d %d %.1f\n",(int(s)+2-bd)/2,bd,s/2);
if(ll!=t)
{
printf("\n");
}
}
return 0;
}