/* poj 1654 Area 多邊形面積題目意思很簡單，但是1000000的point開不了 */ #include<stdio.h> #include<math.h> #include<string.h> const int N=1000000+10; const double eps=1e-8; struct point {double x,y;point(){}point(double a,double b):x(a),y(b){} }; int len,index; char t_s[N]; //基礎函數/ double mo_distance(point p1,point p2) {return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); } double mo_xmult(point p2,point p0,point p1)//p1在p2左返回負，在右邊返回正 {return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); }bool mo_ee(double x,double y) {double ret=x-y;if(ret<0) ret=-ret;if(ret<eps) return 1;return 0; } bool mo_gg(double x,double y) { return x > y + eps;} // x > y bool mo_ll(double x,double y) { return x < y - eps;} // x < y bool mo_ge(double x,double y) { return x > y - eps;} // x >= y bool mo_le(double x,double y) { return x < y + eps;} // x <= y //求多邊形面積 double area_polygon(point shang,point zhe) {point yuan;yuan.x=yuan.y=0;return mo_xmult(zhe,yuan,shang)/2; }/// int getpoint(point shang,point &zhe) {if(t_s[index]=='5')return 0;;char doo=t_s[index];int n=1;while(t_s[index+1]==t_s[index]){++n;++index;}++index;if(doo=='1'){zhe.x=shang.x-n;zhe.y=shang.y-n;}else if(doo=='2'){zhe.x=shang.x;zhe.y=shang.y-n;}else if(doo=='3'){zhe.x=shang.x+n;zhe.y=shang.y-n;}else if(doo=='4'){zhe.x=shang.x-n;zhe.y=shang.y;}else if(doo=='9'){zhe.x=shang.x+n;zhe.y=shang.y+n;}else if(doo=='6'){zhe.x=shang.x+n;zhe.y=shang.y;}else if(doo=='7'){zhe.x=shang.x-n;zhe.y=shang.y+n;}else if(doo=='8'){zhe.x=shang.x;zhe.y=shang.y+n;}return 1; } int main() {int t;scanf("%d",&t);getchar();while(t--){gets(t_s);len=strlen(t_s);index=0;point shang(0,0),zhe;double area=0;while(getpoint(shang,zhe)){area+=area_polygon(shang,zhe);shang=zhe;}if(area<0)area=-area;__int64 ret=(__int64)area,i;for(i=ret;i<=ret+2;++i){if(mo_ee((double)i,area)){printf("%I64d\n",i);break;}else if(mo_ee(0.5+i,area)){printf("%I64d.5\n",i);}}}return 0; }

轉載于:https://www.cnblogs.com/dyllove98/p/3233905.html

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