链接：http://poj.org/problem?id=1127
并查集+线段相交，除了混了一下，没什么难度
View Code
1 #include<stdio.h> 2 #include<string.h> 3 #define N 20 4 #define min(x,y) (x<y?x:y) 5 #define max(x,y) (x>y?x:y) 6 int f[N]; 7 int n; 8 struct point 9 { 10 int x,y; 11 }; 12 struct segment 13 { 14 point p1,p2; 15 }; 16 segment s[N]; 17 void init() 18 { 19 int i; 20 for(i=1;i<=n;i++) 21 f[i]=i; 22 } 23 int find(int i) 24 { 25 int r=i; 26 while(r!=f[r]) 27 r=f[r]; 28 int t; 29 while(i!=f[i]) 30 { 31 t=f[i]; 32 f[i]=r; 33 i=t; 34 } 35 return r; 36 } 37 void uni(int i,int j) 38 { 39 int t1=find(i); 40 int t2=find(j); 41 if(t1^t2) 42 f[t2]=t1; 43 } 44 int mul(point p1,point p2,point p3) 45 { 46 return (p2.x-p1.x)*(p3.y-p2.y)-(p3.x-p2.x)*(p2.y-p1.y); 47 } 48 bool cross(segment s1,segment s2)//线段是否相交 49 { 50 if(min(s1.p1.x,s1.p2.x)>max(s2.p1.x,s2.p2.x)||max(s1.p1.x,s1.p2.x)<min(s2.p1.x,s2.p2.x)) 51 return false; 52 if(min(s1.p1.y,s1.p2.y)>max(s2.p1.y,s2.p2.y)||max(s1.p1.y,s1.p2.y)<min(s2.p1.y,s2.p2.y)) 53 return false; 54 if(mul(s1.p1,s1.p2,s2.p1)*mul(s1.p1,s1.p2,s2.p2)<=0&&mul(s2.p1,s2.p2,s1.p1)*mul(s2.p1,s2.p2,s1.p2)<=0) 55 return true; 56 return false; 57 } 58 int main() 59 { 60 int a,b; 61 int i,j; 62 while(scanf("%d",&n)&&n) 63 { 64 init(); 65 for(i=1;i<=n;i++) 66 scanf("%d%d%d%d",&s[i].p1.x,&s[i].p1.y,&s[i].p2.x,&s[i].p2.y); 67 for(i=1;i<=n;i++) 68 for(j=1;j<=n;j++) 69 { 70 if(i==j) 71 continue; 72 if(cross(s[i],s[j])) 73 uni(i,j); 74 } 75 while(scanf("%d%d",&a,&b)&&(a||b)) 76 { 77 if(find(a)^find(b)) 78 printf("NOT CONNECTED\n"); 79 else 80 printf("CONNECTED\n"); 81 } 82 } 83 return 0; 84 }