為什么注釋掉的地方是錯的？ ?自己的代碼好糟爛.....

直接枚舉點 ?判是否在多邊形內 ?加起來求概率 ? ?求面積的時候代碼寫搓了....? ? ?比不過別人兩行的代碼 ? ?而且到現在還找不到錯.....

#include <iostream> #include <fstream> #include <cstring> #include <cstdio> #include <cmath> #include <cstdlib> #define eps 1e-8 #define _sign(x) ((x)>eps?1:((x)<-eps?2:0)) #define zero(x) (((x)>0?(x):-(x))<eps) #define offset 500 using namespace std; const int N = 100; struct point {double x, y;point(double i, double j){x = i, y = j;}point() {} } p[N];double xmult(point p1,point p2,point p0) {return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } int inside(point q, int n, point *p) {int s[3] = {1, 1, 1};for(int i = 0; i < n && s[1]|s[2]; i++)s[_sign(xmult(p[(i+1)%n], q, p[i]))] = 0;return s[1]|s[2]; } int inside_polygon(point q,int n,point* p) {point q2;int i=0,count;while (i<n)for (count=i=0,q2.x=rand()+offset,q2.y=rand()+offset; i<n; i++){if (zero(xmult(q,p[i],p[(i+1)%n]))&&(p[i].x-q.x)*(p[(i+1)%n].x-q.x)<eps&&(p[i].y-q.y)*(p[(i+1)%n].y-q.y)<eps)return 1;else if (zero(xmult(q,q2,p[i])))break;else if (xmult(q,p[i],q2)*xmult(q,p[(i+1)%n],q2)<-eps&&xmult(p[i],q,p[(i+1)%n])*xmult(p[i],q2,p[(i+1)%n])<-eps)count++;}return count&1; } int main(void) {double x0,y0,xn,yn;int a,b,n;while(scanf("%lf%lf%lf%lf",&x0,&y0,&xn,&yn) == 4){int _x0 = ceil(x0), _y0 = ceil(y0), _xn = floor(xn), _yn = floor(yn); // printf("%d %d %d %d",_x0, _y0, _xn, _yn);scanf("%d%d%d",&n,&a,&b);for(int i = 0; i < n; i++){scanf("%lf%lf",&p[i].x,&p[i].y);}double sum = 0;for(int i = _x0; i <= _xn; i++)for(int j = _y0; j <= _yn; j++){if(inside_polygon(point(i, j), n, p)){double r=(min(i+0.5,xn)-max(i-0.5,x0))*(min(j+0.5,yn)-max(j-0.5,y0));sum += r*(i*a+j*b); // if(i > _x0 && i < _xn && j > _y0 && j < _yn) // sum += (i*a+j*b); // else if(i > _x0 && i < _xn && (j == _y0 || j == _yn)) // { // if(j == _y0) // sum += (i*a+j*b)*((double)((double)_y0-y0+0.5)); // else // sum += (i*a+j*b)*((yn-(double)_yn+0.5)); // } // else if((i == _x0 || i == _xn) && j > _y0 && j < _yn) // { // if(i == _x0) // sum += (i*a+j*b)*(((double)_x0-x0+0.5)); // else // sum += (i*a+j*b)*((xn-(double)_xn+0.5)); // } // else // { // if(i == _x0 && j == _y0) // sum += (i*a+j*b)*(((double)_x0-x0+0.5))*(((double)_y0-y0+0.5)); // else if(i == _x0 && j == _yn) // sum += (i*a+j*b)*(((double)_x0-x0+0.5))*((yn-(double)_yn+0.5)); // else if(i == _xn && j == _y0) // sum += (i*a+j*b)*((xn-(double)_xn+0.5))*(((double)_y0-y0+0.5)); // else // sum += (i*a+j*b)*((xn-(double)_xn+0.5))*((yn-(double)_yn+0.5)); // }}}printf("%.3lf\n",(double)sum/(xn-x0)/(yn-y0));}return 0; }

轉載于:https://www.cnblogs.com/avema/p/3774242.html

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