1 #include <iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<algorithm>  5 #include<stdlib.h>  6 #include<vector>  7 #include<cmath>  8 #include<queue>  9 #include<set> 10 using namespace std; 11 #define N 110 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 const int MAXN=1550; 18 int m,n; 19 double r; 20 int cCnt,curCnt;//此时cCnt为最终切割得到的多边形的顶点数、暂存顶点个数 21 struct point 22 { 23     double x,y; 24     point(double x=0,double y=0):x(x),y(y){} 25 }; 26 struct node 27 { 28     int v,u,w; 29 }ll[N]; 30 point points[MAXN],p[MAXN],q[MAXN];//读入的多边形的顶点（顺时针）、p为存放最终切割得到的多边形顶点的数组、暂存核的顶点 31 void getline(point x,point y,double &a,double &b,double   &c) //两点x、y确定一条直线a、b、c为其系数 32 { 33     a = y.y - x.y; 34     b = x.x - y.x; 35     c = y.x * x.y - x.x * y.y; 36 } 37 void initial() 38 { 39     for(int i = 1; i <= m; ++i)p[i] = points[i]; 40     p[m+1] = p[1]; 41     p[0] = p[m]; 42     cCnt = m;//cCnt为最终切割得到的多边形的顶点数，将其初始化为多边形的顶点的个数 43 } 44 point intersect(point x,point y,double a,double b,double c) //求x、y形成的直线与已知直线a、b、c、的交点 45 { 46     double u = fabs(a * x.x + b * x.y + c); 47     double v = fabs(a * y.x + b * y.y + c); 48     point pt; 49     pt.x=(x.x * v + y.x * u) / (u + v); 50     pt.y=(x.y * v + y.y * u) / (u + v); 51     return  pt; 52 } 53 int dcmp(double x) 54 { 55     if(fabs(x)<eps) return 0; 56     return x<0?-1:1; 57 } 58 void cut(double a,double b ,double c) 59 { 60  61     curCnt = 0; 62     for(int i = 1; i <= cCnt; ++i) 63     { 64         if(a*p[i].x + b*p[i].y + c >-eps)q[++curCnt] = p[i];// c由于精度问题，可能会偏小，所以有些点本应在右侧而没在， 65         //故应该接着判断 66         else 67         { 68             if(a*p[i-1].x + b*p[i-1].y + c > eps) //如果p[i-1]在直线的右侧的话， 69             { 70                 //则将p[i],p[i-1]形成的直线与已知直线的交点作为核的一个顶点(这样的话，由于精度的问题，核的面积可能会有所减少) 71                 q[++curCnt] = intersect(p[i],p[i-1],a,b,c); 72             } 73             if(a*p[i+1].x + b*p[i+1].y + c > eps) //原理同上 74             { 75                 q[++curCnt] = intersect(p[i],p[i+1],a,b,c); 76             } 77         } 78     } 79     for(int i = 1; i <= curCnt; ++i)p[i] = q[i];//将q中暂存的核的顶点转移到p中 80     p[curCnt+1] = q[1]; 81     p[0] = p[curCnt]; 82     cCnt = curCnt; 83 } 84 void solve(int k) 85 { 86     //注意：默认点是顺时针，如果题目不是顺时针，规整化方向 87     initial(); 88     for(int i = 1; i <= n; ++i) 89     { 90         if(i==k) continue; 91         double a,b,c; 92         a = (ll[k].u-ll[i].u)*1.0/(ll[k].u*ll[i].u); 93         b = (ll[k].w-ll[i].w)*1.0/(ll[k].w*ll[i].w); 94         c = (ll[k].v-ll[i].v)*1.0/(ll[k].v*ll[i].v); 95         if(dcmp(a)==0&&dcmp(b)==0&&dcmp(c)<=0) 96         { 97             puts("No"); 98             return ; 99         }100         //getline(points[i],points[i+1],a,b,c);101         cut(a,b,c);102     }103     /*104       如果要向内推进r，用该部分代替上个函数105       for(int i = 1; i <= m; ++i){106           Point ta, tb, tt;107           tt.x = points[i+1].y - points[i].y;108           tt.y = points[i].x - points[i+1].x;109           double k = r / sqrt(tt.x * tt.x + tt.y * tt.y);110           tt.x = tt.x * k;111           tt.y = tt.y * k;112           ta.x = points[i].x + tt.x;113           ta.y = points[i].y + tt.y;114           tb.x = points[i+1].x + tt.x;115           tb.y = points[i+1].y + tt.y;116           double a,b,c;117           getline(ta,tb,a,b,c);118           cut(a,b,c);119       }*/120     //多边形核的面积121     double area = 0;122     for(int i = 1; i <= cCnt; ++i)123         area += p[i].x * p[i + 1].y - p[i + 1].x * p[i].y;124     area = fabs(area / 2.0);125     if(dcmp(area)>0)126     printf("Yes\n");127     else128     puts("No");129 130 }131 /*void GuiZhengHua(){132      //规整化方向，逆时针变顺时针，顺时针变逆时针133     for(int i = 1; i < (m+1)/2; i ++)134       swap(points[i], points[m-i]);135 }*/136 int main()137 {138     points[1] = point(0,0);139     points[2] = point(INF,0);140     points[3] = point(INF,INF);141     points[4] = point(0,INF);142     points[5] = points[1];143     m = 4;144     int i;145     while(scanf("%d",&n)!=EOF)146     {147         for(i = 1; i <=n ;i++)148         scanf("%d%d%d",&ll[i].v,&ll[i].u,&ll[i].w);149         for(i = 1;i <= n ;i++)150         solve(i);151     }152     return 0;153 }