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