叉积求面积
hdu2036（模版题）:
http://acm.hdu.edu.cn/showproblem.php?pid=2036
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1 #include <iostream> 2 #include<cstdio> 3 #include<string.h> 4 #include<math.h> 5 using namespace std; 6 struct node 7 { 8 int x,y; 9 }q[101]; 10 double cross(node a,node b) 11 { 12 return a.x*b.y-a.y*b.x; 13 } 14 int main() 15 { 16 int i,j,k,n; 17 while(scanf("%d",&n)&&n) 18 { 19 double s = 0; 20 for(i = 1; i <= n ; i++) 21 scanf("%d%d",&q[i].x,&q[i].y); 22 for(i = 1 ; i <= n ; i++) 23 { 24 if(i<n) 25 s+=cross(q[i],q[i+1]); 26 else 27 s+=cross(q[n],q[1]); 28 } 29 if(s<0) 30 s = -s; 31 s = s/2; 32 printf("%.1lf\n",s); 33 } 34 return 0; 35 }
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=41
求怎样旋转这个半圆能包围最多的点
1.到圆心的距离大于半径的点直接排除。
2.以圆心和任意一点确定一有向线段作为半径位置，分别计数该有向线段左边点的个数(nl)和右边点的个数(nr)。
重复步骤2直到所有点都被枚举
过。
4.枚举过程中出现的最大的nl或
nr就是所求的结果。
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1 #include <iostream> 2 #include<cstdio> 3 #include<string.h> 4 using namespace std; 5 struct node 6 { 7 int x,y; 8 }no[200]; 9 int cross(int x1,int y1,int x2,int y2) 10 { 11 return x1*y2-x2*y1; 12 } 13 int main() 14 { 15 int i,j,k,n,m,x,y,g; 16 double r; 17 while(scanf("%d%d%lf",&n,&m,&r)!=EOF) 18 { 19 if(r<0) 20 break; 21 scanf("%d",&k); 22 g = 0; 23 for(i = 1; i <= k ;i++) 24 { 25 scanf("%d%d",&x,&y); 26 if((x-n)*(x-n)+(y-m)*(y-m)<=r*r)//排除到圆心距离大于半径的点 27 { 28 g++; 29 no[g].x = x; 30 no[g].y = y; 31 } 32 } 33 int max = 0; 34 for(i = 1; i <= g ; i++) 35 { 36 int numl = 0,numr = 0; 37 for(j = 1; j <= g ; j++) 38 { 39 int d = cross(no[i].x-n,no[i].y-m,no[j].x-n,no[j].y-m);//点乘判断是哪边的点 40 if(d==0) 41 { 42 numl++; 43 numr++; 44 } 45 if(d>0) 46 numl++; 47 if(d<0) 48 numr++; 49 } 50 if(numl>max) 51 max = numl; 52 if(numr>max) 53 max = numr; 54 } 55 printf("%d\n",max); 56 } 57 return 0; 58 }
规范相交题目 利用差乘判断相交 不用考虑重点问题
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1 #include <iostream> 2 #include<cstdio> 3 #include<string.h> 4 using namespace std; 5 double eps=0.0000000001; 6 struct node 7 { 8 double x1,x2,y1,y2; 9 }; 10 double cross(double x1,double x2,double y1,double y2) 11 { 12 return x1*y2-x2*y1; 13 } 14 int main() 15 { 16 int i,j,k,n,m; 17 node q[101]; 18 while(scanf("%d",&n)&&n) 19 { 20 int num = 0; 21 for(i = 1; i <= n ; i++) 22 { 23 scanf("%lf%lf%lf%lf",&q[i].x1,&q[i].y1,&q[i].x2,&q[i].y2); 24 } 25 for(i = 1; i < n; i++) 26 for(j = i+1 ; j <= n ; j++) 27 { 28 double d1 = cross(q[i].x1-q[i].x2,q[i].y1-q[i].y2,q[i].x1-q[j].x1,q[i].y1-q[j].y1); 29 double d2 = cross(q[i].x1-q[i].x2,q[i].y1-q[i].y2,q[i].x1-q[j].x2,q[i].y1-q[j].y2); 30 double d3 = cross(q[j].x1-q[j].x2,q[j].y1-q[j].y2,q[j].x1-q[i].x2,q[j].y1-q[i].y2); 31 double d4 = cross(q[j].x1-q[j].x2,q[j].y1-q[j].y2,q[j].x1-q[i].x1,q[j].y1-q[i].y1); 32 if(d1*d2<eps&&d3*d4<eps) 33 num++; 34 } 35 printf("%d\n",num); 36 } 37 return 0; 38 }
二分查找+点乘判断左右
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1 #include <iostream> 2 #include<cstdio> 3 #include<string.h> 4 using namespace std; 5 struct node 6 { 7 int x1,x2; 8 }q[5001]; 9 int cross(node a,node b) 10 { 11 if(a.x1*b.x2-a.x2*b.x1>0) 12 return 1; 13 else 14 return 0; 15 } 16 int x1,x2,n,y1,y2; 17 int judge(int x,int y) 18 { 19 int l =0,r = n+1,mid,k; 20 node a,b; 21 while(l<r) 22 { 23 mid = (l+r)/2; 24 if(mid==l) 25 return mid; 26 a.x1 = q[mid].x1-q[mid].x2; 27 a.x2 = y1-y2; 28 b.x1 = x-q[mid].x2; 29 b.x2 = y-y2; 30 if(cross(a,b)) 31 r = mid; 32 else 33 l = mid; 34 } 35 return mid; 36 } 37 int num[5010]; 38 int main() 39 { 40 int i,j,m,t,x,y; 41 while(scanf("%d",&n)&&n) 42 { 43 memset(num,0,sizeof(num)); 44 scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2); 45 for(i = 1; i <= n ; i++) 46 scanf("%d%d",&q[i].x1,&q[i].x2); 47 while(m--) 48 { 49 scanf("%d%d",&x,&y); 50 int k = judge(x,y); 51 num[k]++; 52 } 53 for(i = 0 ; i <= n ; i++) 54 printf("%d: %d\n",i,num[i]); 55 puts(""); 56 } 57 return 0; 58 }