链接：http://poj.org/problem?id=1265
Area
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 4969 | Accepted: 2231 |
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.
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
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Pick定理的证明推荐个网址：episte.math.ntu.edu.tw/articles/sm/sm_25_10_1/page2.html
Area=i + b/2 - 1
i为内点 b为边上点
还有点在多边形边上运用GCD的证明，确实不会，希望大神看到给我解释一下
超时代码：
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm> #define eps 1e-8
#define MAXX 210
using namespace std; typedef struct point
{
double x;
double y;
}point;
typedef struct line
{
point st;
point ed;
}line; bool dy(double x,double y)
{
return x>y+eps;
}
bool xy(double x,double y)
{
return x<y-eps;
}
bool dyd(double x,double y)
{
return x>y-eps;
}
bool xyd(double x,double y)
{
return x<y+eps;
}
bool dd(double x,double y)
{
return fabs(x-y)<eps;
} double crossProduct(point a,point b,point c)//ac -> ab
{
return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
}
double dist(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
} bool onSegment_1(point a,point b,point c)
{
double maxx=max(a.x,b.x);
double minx=min(a.x,b.x);
double maxy=max(a.y,b.y);
double miny=min(a.y,b.y); if(dd(crossProduct(a,b,c),0.0) && xyd(c.x,maxx) && dyd(c.x,minx)
&& xyd(c.y,maxy) && dyd(c.y,miny))
return true;
return false;
} bool segIntersect_1(point p1,point p2,point p3,point p4)
{
double d1=crossProduct(p3,p4,p1);
double d2=crossProduct(p3,p4,p2);
double d3=crossProduct(p1,p2,p3);
double d4=crossProduct(p1,p2,p4); if(xy(d1*d2,0.0) && xy(d3*d4,0.0))
return true;
if(dd(d1,0.0) && onSegment_1(p3,p4,p1))
return true;
if(dd(d2,0.0) && onSegment_1(p3,p4,p2))
return true;
if(dd(d3,0.0) && onSegment_1(p1,p2,p3))
return true;
if(dd(d4,0.0) && onSegment_1(p1,p2,p4))
return true;
return false;
} point p[MAXX];
line li[MAXX];
int n; bool inPolygon_1(point pot)
{
int count=;
line l;
l.st=pot;
l.ed.x=1e10;
l.ed.y=pot.y;
p[n]=p[];
for(int i=; i<n; i++)
{
if( onSegment_1(p[i],p[i+],pot ))
return true;
if(!dd(p[i].y,p[i+].y))
{
int tmp=-;
if(onSegment_1(l.st,l.ed,p[i]))
tmp=i;
else if(onSegment_1(l.st,l.ed,p[i+]))
tmp=i+;
if(tmp != - && dd(p[tmp].y,max(p[i].y,p[i+].y)))
count++;
else if(tmp == - && segIntersect_1(p[i],p[i+],l.st,l.ed))
count++;
}
}
if(count % ==)
return true;
return false;
} bool inPolygon_2(point pot)
{
int count=;
line l;
l.st=pot;
l.ed.x=1e10;
l.ed.y=pot.y;
p[n]=p[];
for(int i=; i<n; i++)
{
if( onSegment_1(p[i],p[i+],pot ))
return false;
if(!dd(p[i].y,p[i+].y))
{
int tmp=-;
if(onSegment_1(l.st,l.ed,p[i]))
tmp=i;
else if(onSegment_1(l.st,l.ed,p[i+]))
tmp=i+;
if(tmp != - && dd(p[tmp].y,max(p[i].y,p[i+].y)))
count++;
else if(tmp == - && segIntersect_1(p[i],p[i+],l.st,l.ed))
count++;
}
}
if(count % ==)
return true;
return false;
} double Area(int n)
{
if(n<)return ;
int i;
double ret=0.0;
for(i=; i<=n; i++)
{
ret+=(crossProduct(p[],p[i-],p[i]));
}
return fabs(ret)/2.0;
} int main()
{
int m,i,j;
int ttmp=;
scanf("%d",&m);
while(m--)
{
scanf("%d",&n);
p[].x=;p[].y=;
double x,y;
double maxx=-,maxy=-,minx=,miny=;
for(i=; i<=n; i++)
{
scanf("%lf%lf",&x,&y);
p[i].x=p[i-].x+x;
p[i].y=p[i-].y+y;//printf("%lf %lf^^",p[i].x,p[i].y);
}
for(i=; i<=n; i++)
{
maxx=maxx>p[i].x?maxx:p[i].x;
maxy=maxy>p[i].y?maxy:p[i].y;
minx=minx<p[i].x?minx:p[i].x;
miny=miny<p[i].y?miny:p[i].y;
}//printf("%lf %lf^^%lf %lf**",minx,maxx,miny,maxy);
int in=,edge=,sum=;
point cas;
for(i=minx; i<=maxx; i++)
{
for(j=miny; j<=maxy; j++)
{
cas.x=i;cas.y=j;
if(inPolygon_1(cas))
{
sum++;
}
if(inPolygon_2(cas))
{
in++;
}
}
}
double area=Area(n);
edge=sum-in;
printf("Scenario #%d:\n",ttmp++);
printf("%d %d %.1lf\n",in,edge,area);
}
return ;
}
Pick定理运用
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm> #define eps 1e-8
#define MAXX 210 typedef struct point
{
double x;
double y;
}point; int gcd(int a,int b)
{
return b ? gcd(b,a%b) : a;
} double crossProduct(point a,point b,point c)
{
return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
} point p[MAXX];
int onBorder(int n)
{
int sum=;
for(int i=; i<n; i++)
{
sum+=gcd(abs((int)(p[i].x-p[i+].x)),abs((int)(p[i].y-p[i+].y)));
}
return sum;
} double area(int n)
{
double ans=0.0;
for(int i=; i<=n; i++)
{
ans+=crossProduct(p[],p[i-],p[i]);
}
return fabs(ans)/2.0;
} int main()
{
int n,m,i,j;
scanf("%d",&n);
int cas=;
while(n--)
{
scanf("%d",&m);
p[].x=;p[].y=;
double x,y;
for(i=; i<=m; i++)
{
scanf("%lf%lf",&x,&y);
p[i].x=p[i-].x+x;
p[i].y=p[i-].y+y;
}
double are=area(m);
int edge=onBorder(m);
printf("Scenario #%d:\n",cas++);
printf("%d %d %.1lf\n\n",(int)are+-edge/,edge,are);
}
return ;
}