poj 1265 Area
来源:互联网 发布:广联达算量软件多少钱 编辑:程序博客网 时间:2024/06/03 19:09
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. 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 dx 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 Scenario #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
241 00 1-1 00 -175 01 3-2 2-1 00 -3-3 10 -3
Sample Output
Scenario #1:0 4 1.0Scenario #2:12 16 19.0
【题意】
给你很多个点,让你求出这些点围成的多边行的内部点的数量,边上点的数量,多边形的面积。但是需要注意的是,这里给的点的顺序是逆时针,还有就是给出的并不是点的坐标而是相对于前一个点的坐标变化量也就是dxdy。
【分析】
多边形的面积就是拆成三角形的面积然后累加得到的,好吧说实话,我是百度的,用的叉积。然后还有一个就是pick公式,说实话也是第一次见,以后得记住了。
【代码】
#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>using namespace std;int gcd(int a,int b){ return b?gcd(b,a%b):a;}typedef struct{ int x,y;}P;P poi[105];int main(){ int T; scanf("%d",&T); for(int z=1;z<=T;z++) { printf("Scenario #%d:\n",z); int num; scanf("%d",&num); int edge=0; int area=0; poi[0].x=poi[0].y=0; for(int i=1;i<=num;i++) { scanf("%d %d",&poi[i].x,&poi[i].y); edge+=gcd(abs(poi[i].x),abs(poi[i].y)); poi[i].x+=poi[i-1].x; poi[i].y+=poi[i-1].y; area+=poi[i-1].x*poi[i].y-poi[i].x*poi[i-1].y; } printf("%d %d ",area/2+1-edge/2,edge); printf(area&1?"%d.5\n":"%d.0\n",area/2); puts(""); } return 0;}
阅读全文
0 0
- Poj 1265 Area
- poj 1265 Area
- POJ 1265 Area
- poj 1265 Area
- POJ 1265 Area
- POJ 1265 Area
- POJ 1265 Area
- POJ 1265 Area
- poj 1265 Area
- POJ 1265:Area
- POJ 1265 Area
- POJ 1265 Area
- poj 1265 Area
- poj-1265 Area
- poj 1265 Area
- POJ 1265 Area
- POJ 1265 Area
- poj 1265 Area
- JQuery--属性选择器
- 使用递归来解决在单链表里删除第一节点不是数字‘2’
- Ctrl+C, Ctrl+Z, Ctrl+D
- Web架构(一)PHP前端请求Java接口
- Django开发中时间的时区的处理
- poj 1265 Area
- Unity3d 技巧(8) -PlayMaker 插件自定义扩展 不受局限
- CTF实验吧-简单的sql注入3【sqlmap直接跑】
- 学习笔记_02CNN
- ActionBar图标文字隐藏:
- 外卖聚合服务性能测试经验总结
- JavaScript NaN 属性
- Python3.x
- C++复合类型:指针和引用