第一道计算几何题。。。。
试用了下两种删除vector中元素的方法。
第一种使用：（每次erase一个元素后都会指向下一个元素位置）
1 for(int j=0;j<v.size();) 2 { 3 if(intersect_in(lin[i],lin[v[j]])) 4 v.erase(v.begin()+j); 5 else 6 j++; 7 }
View Code
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 8 using namespace std; 9 10 #define eps 1e-8 11 #define PI acos(-1.0)//3.14159265358979323846 12 //判断一个数是否为0,是则返回true,否则返回false 13 #define zero(x)(((x)>0?(x):-(x))<eps) 14 //返回一个数的符号，正数返回1，负数返回2，否则返回0 15 #define _sign(x)((x)>eps?1:((x)<-eps?2:0)) 16 17 const int maxn = 1e5+5; 18 struct point 19 { 20 double x,y; 21 }; 22 struct line 23 { 24 point a,b; 25 line(){} 26 line(double x1,double y1,double x2,double y2) 27 { 28 a.x=x1; 29 a.y=y1; 30 b.x=x2; 31 b.y=y2; 32 } 33 }; 34 line lin[maxn]; 35 double xmult(point p1,point p2,point p0) 36 { 37 return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); 38 } 39 //计算dotproduct(P1-P0).(P2-P0) 40 double dmult(point p1,point p2,point p0) 41 { 42 return(p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y); 43 } 44 //判三点共线 45 int dots_inline(point p1,point p2,point p3) 46 { 47 return zero(xmult(p1,p2,p3)); 48 } 49 //判点是否在线段上,包括端点 50 int dot_online_in(point p,line l) 51 { 52 return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps; 53 } 54 //判两点在线段同侧,点在线段上返回0 55 int same_side(point p1,point p2,line l) 56 { 57 return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)>eps; 58 } 59 60 int intersect_in(line u,line v) 61 { 62 if(!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b))//规范相交 63 return!same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u); 64 return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u); 65 } 66 67 int main() 68 { 69 int n; 70 double x1,x2,y1,y2; 71 while(scanf("%d",&n)) 72 { 73 if(!n) break; 74 vector<int>v; 75 for(int i=1;i<=n;i++) 76 { 77 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 78 lin[i]=line(x1,y1,x2,y2); 79 for(int j=0;j<v.size();) 80 { 81 if(intersect_in(lin[i],lin[v[j]])) 82 v.erase(v.begin()+j); 83 else 84 j++; 85 } 86 v.push_back(i); 87 } 88 printf("Top sticks:"); 89 for(int i=0;i<v.size()-1;i++) 90 printf(" %d,",v[i]); 91 printf(" %d.\n",v[v.size()-1]); 92 } 93 return 0; 94 }
第二种使用remove_if()。
1 line tmp; 2 bool ok(line l) 3 { 4 if(intersect_in(tmp,l)) 5 return true; 6 else 7 return false; 8 } 9 10 v.erase(remove_if(v.begin(),v.end(),ok),v.end());
View Code
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 8 using namespace std; 9 10 #define eps 1e-8 11 #define PI acos(-1.0)//3.14159265358979323846 12 //判断一个数是否为0,是则返回true,否则返回false 13 #define zero(x)(((x)>0?(x):-(x))<eps) 14 //返回一个数的符号，正数返回1，负数返回2，否则返回0 15 #define _sign(x)((x)>eps?1:((x)<-eps?2:0)) 16 17 struct point 18 { 19 double x,y; 20 }; 21 struct line 22 { 23 point a,b; 24 int id; 25 line(){} 26 line(double x1,double y1,double x2,double y2,int i) 27 { 28 a.x=x1; 29 a.y=y1; 30 b.x=x2; 31 b.y=y2; 32 id=i; 33 } 34 }; 35 double xmult(point p1,point p2,point p0) 36 { 37 return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); 38 } 39 //计算dotproduct(P1-P0).(P2-P0) 40 double dmult(point p1,point p2,point p0) 41 { 42 return(p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y); 43 } 44 //判三点共线 45 int dots_inline(point p1,point p2,point p3) 46 { 47 return zero(xmult(p1,p2,p3)); 48 } 49 //判点是否在线段上,包括端点 50 int dot_online_in(point p,line l) 51 { 52 return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps; 53 } 54 //判两点在线段同侧,点在线段上返回0 55 int same_side(point p1,point p2,line l) 56 { 57 return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)>eps; 58 } 59 60 int intersect_in(line u,line v) 61 { 62 if(!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b))//规范相交 63 return!same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u); 64 return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u); 65 } 66 67 line tmp; 68 bool ok(line l) 69 { 70 if(intersect_in(tmp,l)) 71 return true; 72 else 73 return false; 74 } 75 76 int main() 77 { 78 int n; 79 double x1,x2,y1,y2; 80 while(scanf("%d",&n)) 81 { 82 if(!n) break; 83 vector<line>v; 84 for(int i=1;i<=n;i++) 85 { 86 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 87 tmp=line(x1,y1,x2,y2,i); 88 v.erase(remove_if(v.begin(),v.end(),ok),v.end()); 89 /*for(int j=0;j<v.size();) 90 { 91 if(intersect_in(lin[i],lin[v[j]])) 92 v.erase(v.begin()+j); 93 else 94 j++; 95 }*/ 96 v.push_back(tmp); 97 98 } 99 printf("Top sticks:"); 100 for(int i=0;i<v.size()-1;i++) 101 printf(" %d,",v[i].id); 102 printf(" %d.\n",v[v.size()-1].id); 103 } 104 return 0; 105 }