枚举两点 若不和任何线段相交 建边为dis（i,j) floyd求最短路
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 100 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 struct point 18 { 19 double x,y; 20 point(double x=0,double y=0):x(x),y(y){} 21 }p[N]; 22 struct line 23 { 24 point u,v; 25 }li[N]; 26 double w[N][N]; 27 typedef point pointt; 28 pointt operator - (point a,point b) 29 { 30 return pointt(a.x-b.x,a.y-b.y); 31 } 32 int dcmp(double x) 33 { 34 if(fabs(x)<eps) return 0; 35 return x<0?-1:1; 36 } 37 double dis(point a) 38 { 39 return sqrt(a.x*a.x+a.y*a.y); 40 } 41 double cross(point a,point b) 42 { 43 return a.x*b.y-a.y*b.x; 44 } 45 bool segprointer(point a1,point a2,point b1,point b2) 46 { 47 double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1), 48 c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1); 49 return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0; 50 } 51 int main() 52 { 53 int n,i,j,k; 54 while(scanf("%d",&n)!=EOF) 55 { 56 if(n==-1) break; 57 int g = 0; 58 for(i = 1; i <= 100 ; i++) 59 { 60 for(j = 1; j<= 100 ; j++) 61 w[i][j] = INF; 62 w[i][i] = 0; 63 } 64 int o = 0; 65 for(i = 1; i <= n ;i++) 66 { 67 double k; 68 scanf("%lf",&k); 69 for(j = 1; j <= 4; j++) 70 { 71 p[++g].x = k; 72 scanf("%lf",&p[g].y); 73 } 74 point pp = point(k,0); 75 li[++o].u = pp; 76 li[o].v = p[g-3]; 77 li[++o].u = p[g-2]; 78 li[o].v = p[g-1]; 79 li[++o].u = p[g]; 80 pp = point(k,10); 81 li[o].v = pp; 82 } 83 p[g+1] = point(0,5); 84 p[g+2] = point(10,5); 85 //printf("%d ",segprointer(p[g+1],p[g+2],li[5].u,li[5].v)); 86 for(i = 1; i <= g+2; i++) 87 for(j = i+1; j <= g+2; j++) 88 { 89 if(i==j) continue; 90 for(k = 1; k <= o ; k++) 91 { 92 if(segprointer(p[i],p[j],li[k].u,li[k].v))break; 93 94 } 95 if(k>o) 96 w[i][j] = w[j][i] = dis(p[i]-p[j]); 97 //printf("%.2f %.2f %.2f %.2f %.2f ",p[i].x,p[i].y,p[j].x,p[j].y,w[i][j]); 98 } 99 for(i = 1; i <= g+2 ; i++) 100 for(j = 1; j <=g+2 ;j++) 101 for(k = 1; k <= g+2 ; k++) 102 w[j][k] = min(w[j][i]+w[i][k],w[j][k]); 103 printf("%.2f ",w[g+1][g+2]); 104 } 105 return 0; 106 }