标签:图论
input | output |
---|---|
4 2 0 0 1 0 1 1 0 1 1 3 4 2 |
YES |
4 3 0 0 1 0 1 1 0 1 1 2 2 1 3 4 |
NO |
3 2 0 0 1 0 1 1 1 3 3 2 |
YES |
#include <stdio.h> #include <iostream> #include <math.h> #include <algorithm> using namespace std; struct EDG { int u,v; }; struct point { double x,y; }; point p[305]; double fan(double x,double y) { return x>y?x:y; } double fin(double c,double d) { return c<d?c:d; } double cnt(point a,point b) { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } int is(point a,point b,point c,point d)//判断两条线段是不是连通 { if(a.x==b.x&&c.x==d.x) { return 0; } if(a.x==b.x&&c.x!=d.x) { double m1=a.x; double m2=(a.x-c.x)*(d.y-c.y)/(d.x-c.x)+c.y; if(m1<=fan(a.x,b.x)&&m1>=fin(a.x,b.x)&&m2>=fin(a.y,b.y)&&m2<=fan(a.y,b.y)&&m1<=fan(c.x,d.x)&&m1>=fin(c.x,d.x)&&m2>=fin(c.y,d.y)&&m2<=fan(c.y,d.y)) return 1; } if(c.x==d.x&&a.x!=b.x) { double m1=c.x; double m2=a.y+(b.y-a.y)*(c.x-a.x)/(b.x-a.x); if(m1<=fan(a.x,b.x)&&m1>=fin(a.x,b.x)&&m2>=fin(a.y,b.y)&&m2<=fan(a.y,b.y)&&m1<=fan(c.x,d.x)&&m1>=fin(c.x,d.x)&&m2>=fin(c.y,d.y)&&m2<=fan(c.y,d.y)) return 1; } double k1=(b.y-a.y)/(b.x-a.x); double k2=(d.y-c.y)/(d.x-c.x); double m1,m2,x,y; if(k1==k2) return 0; else { m1=a.y-k1*a.x; m2=c.y-k2*c.x; x=(m1-m2)/(k2-k1); y=k1*x+m1; if(x<=fan(a.x,b.x)&&x>=fin(a.x,b.x)&&y>=fin(a.y,b.y)&&y<=fan(a.y,b.y)&&x<=fan(c.x,d.x)&&x>=fin(c.x,d.x)&&y>=fin(c.y,d.y)&&y<=fan(c.y,d.y)) return 1; } return 0; } int father[305]; int findroot(int x) { if(x!=father[x]) father[x]=findroot(father[x]); return father[x]; } void setroot(int x,int y) { x=findroot(x); y=findroot(y); father[x]=y; } int main() { int cas = 1; int n,m,i,j; EDG edg[305]; while(~scanf("%d%d",&n,&m)) { for(int i=0;i<=n;i++) father[i]=i; int cnt = 0; for(i =1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y); for( i=1;i<=m;i++) scanf("%d%d",&edg[i].u,&edg[i].v); for(i = 1;i<=m;i++) { setroot(edg[i].u,edg[i].v); for(j =1;j<i;j++) { if(is(p[edg[i].u],p[edg[i].v],p[edg[j].u],p[edg[j].v])) { setroot(findroot(edg[i].v),findroot(edg[j].v)); } } } for(i=1;i<=n;i++) if(father[i]==i) cnt++; printf("%s\n",cnt>1?"NO":"YES"); } return 0; }
BNUOJ33566 Cycling Roads(并查集+判断两线段相交)
标签:图论
原文地址:http://blog.csdn.net/u010372095/article/details/45133157