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poj 1265 Area (Pick定理+求面积)
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.
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
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
Source
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超时代码：
1 #include <stdio.h> 2 #include <math.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include <iostream> 6 #include <algorithm> 7 8 #define eps 1e-8 9 #define MAXX 210 10 using namespace std; 11 12 typedef struct point 13 { 14 double x; 15 double y; 16 }point; 17 typedef struct line 18 { 19 point st; 20 point ed; 21 }line; 22 23 bool dy(double x,double y) 24 { 25 return x>y+eps; 26 } 27 bool xy(double x,double y) 28 { 29 return x<y-eps; 30 } 31 bool dyd(double x,double y) 32 { 33 return x>y-eps; 34 } 35 bool xyd(double x,double y) 36 { 37 return x<y+eps; 38 } 39 bool dd(double x,double y) 40 { 41 return fabs(x-y)<eps; 42 } 43 44 double crossProduct(point a,point b,point c)//ac -> ab 45 { 46 return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x); 47 } 48 double dist(point a,point b) 49 { 50 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 51 } 52 53 bool onSegment_1(point a,point b,point c) 54 { 55 double maxx=max(a.x,b.x); 56 double minx=min(a.x,b.x); 57 double maxy=max(a.y,b.y); 58 double miny=min(a.y,b.y); 59 60 if(dd(crossProduct(a,b,c),0.0) && xyd(c.x,maxx) && dyd(c.x,minx) 61 && xyd(c.y,maxy) && dyd(c.y,miny)) 62 return true; 63 return false; 64 } 65 66 bool segIntersect_1(point p1,point p2,point p3,point p4) 67 { 68 double d1=crossProduct(p3,p4,p1); 69 double d2=crossProduct(p3,p4,p2); 70 double d3=crossProduct(p1,p2,p3); 71 double d4=crossProduct(p1,p2,p4); 72 73 if(xy(d1*d2,0.0) && xy(d3*d4,0.0)) 74 return true; 75 if(dd(d1,0.0) && onSegment_1(p3,p4,p1)) 76 return true; 77 if(dd(d2,0.0) && onSegment_1(p3,p4,p2)) 78 return true; 79 if(dd(d3,0.0) && onSegment_1(p1,p2,p3)) 80 return true; 81 if(dd(d4,0.0) && onSegment_1(p1,p2,p4)) 82 return true; 83 return false; 84 } 85 86 point p[MAXX]; 87 line li[MAXX]; 88 int n; 89 90 bool inPolygon_1(point pot) 91 { 92 int count=0; 93 line l; 94 l.st=pot; 95 l.ed.x=1e10; 96 l.ed.y=pot.y; 97 p[n]=p[0]; 98 for(int i=0; i<n; i++) 99 {100 if( onSegment_1(p[i],p[i+1],pot ))101 return true;102 if(!dd(p[i].y,p[i+1].y))103 {104 int tmp=-1;105 if(onSegment_1(l.st,l.ed,p[i]))106 tmp=i;107 else if(onSegment_1(l.st,l.ed,p[i+1]))108 tmp=i+1;109 if(tmp != -1 && dd(p[tmp].y,max(p[i].y,p[i+1].y)))110 count++;111 else if(tmp == -1 && segIntersect_1(p[i],p[i+1],l.st,l.ed))112 count++;113 }114 }115 if(count % 2 ==1)116 return true;117 return false;118 }119 120 bool inPolygon_2(point pot)121 {122 int count=0;123 line l;124 l.st=pot;125 l.ed.x=1e10;126 l.ed.y=pot.y;127 p[n]=p[0];128 for(int i=0; i<n; i++)129 {130 if( onSegment_1(p[i],p[i+1],pot ))131 return false;132 if(!dd(p[i].y,p[i+1].y))133 {134 int tmp=-1;135 if(onSegment_1(l.st,l.ed,p[i]))136 tmp=i;137 else if(onSegment_1(l.st,l.ed,p[i+1]))138 tmp=i+1;139 if(tmp != -1 && dd(p[tmp].y,max(p[i].y,p[i+1].y)))140 count++;141 else if(tmp == -1 && segIntersect_1(p[i],p[i+1],l.st,l.ed))142 count++;143 }144 }145 if(count % 2 ==1)146 return true;147 return false;148 }149 150 double Area(int n)151 {152 if(n<3)return 0;153 int i;154 double ret=0.0;155 for(i=2; i<=n; i++)156 {157 ret+=(crossProduct(p[0],p[i-1],p[i]));158 }159 return fabs(ret)/2.0;160 }161 162 int main()163 {164 int m,i,j;165 int ttmp=1;166 scanf("%d",&m);167 while(m--)168 {169 scanf("%d",&n);170 p[0].x=0;p[0].y=0;171 double x,y;172 double maxx=-100,maxy=-100,minx=100,miny=100;173 for(i=1; i<=n; i++)174 {175 scanf("%lf%lf",&x,&y);176 p[i].x=p[i-1].x+x;177 p[i].y=p[i-1].y+y;//printf("%lf %lf^^",p[i].x,p[i].y);178 }179 for(i=0; i<=n; i++)180 {181 maxx=maxx>p[i].x?maxx:p[i].x;182 maxy=maxy>p[i].y?maxy:p[i].y;183 minx=minx<p[i].x?minx:p[i].x;184 miny=miny<p[i].y?miny:p[i].y;185 }//printf("%lf %lf^^%lf %lf**",minx,maxx,miny,maxy);186 int in=0,edge=0,sum=0;187 point cas;188 for(i=minx; i<=maxx; i++)189 {190 for(j=miny; j<=maxy; j++)191 {192 cas.x=i;cas.y=j;193 if(inPolygon_1(cas))194 {195 sum++;196 }197 if(inPolygon_2(cas))198 {199 in++;200 }201 }202 }203 double area=Area(n);204 edge=sum-in;205 printf("Scenario #%d:\n",ttmp++);206 printf("%d %d %.1lf\n",in,edge,area);207 }208 return 0;209 }
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