Cycling Roads URAL1966 并查集

1966. Cycling Roads

Time limit: 1.0 second
Memory limit: 64 MB

When Vova was in Shenzhen, he rented a bike and spent most of the time cycling around the city. Vova was approaching one of the city parks when he noticed the park plan hanging opposite the central entrance. The plan had several marble statues marked on it. One of such statues stood right there, by the park entrance. Vova wanted to ride in the park on the bike and take photos of all statues. The park territory has multiple bidirectional cycling roads. Each cycling road starts and ends at a marble statue and can be represented as a segment on the plane. If two cycling roads share a common point, then Vova can turn on this point from one road to the other. If the statue stands right on the road, it doesn't interfere with the traffic in any way and can be photoed from the road.

Can Vova get to all statues in the park riding his bike along cycling roads only?

Input

The first line contains integers n and m that are the number of statues and cycling roads in the park (1 ≤ m < n ≤ 200). Then n lines follow, each of them contains the coordinates of one statue on the park plan. The coordinates are integers, their absolute values don't exceed 30 000. Any two statues have distinct coordinates. Each of the following m lines contains two distinct integers from 1 to n that are the numbers of the statues that have a cycling road between them.

Output

Print “YES” if Vova can get from the park entrance to all the park statues, moving along cycling roads only, and “NO” otherwise.

Samples

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

Problem Source: Open Ural FU Personal Contest 2013

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具体自己写：

int main() 
{
    //freopen("in.txt","r",stdin);
    int n,m;
    scanf("%d%d",&n,&m);
    repf(i,0,n)
        p[i] = i;
    
    repf(i,1,n)
        P[i] = read_point();
    
    repf(i,1,m)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        e[i] = edge(a,b);
        int A = find(a);
        int B = find(b);
        if(A != B)
            p[B] = A;
    }
    
    int num = 0;
    
    repf(i,1,n)
        repf(j,1,m)
        {
            if(OnSegment(P[i],P[e[j].a],P[e[j].b]))
            {
                int A = find(i);
                int B = find(e[j].b);
                int C = find(e[j].a);
                p[B] = A;
                p[C] = A;
            }
        }
    repf(i,1,m)
        repf(j,i+1,m)
        {
            int a = e[i].a;
            int b = e[i].b;
            int c = e[j].a;
            int d = e[j].b; 
            if(SegmentProperIntersection(P[a],P[b],P[c],P[d]))
            {
                int A = find(a);
                int B = find(b);
                int C = find(c);
                int D = find(d);
                p[B] = A;
                p[C] = A;
                p[D] = A;
            }
        }
    
    repf(i,1,n)
        if(p[i] == i)
            num++;
    if(num > 1)
        cout<<"NO"<<endl;
    else
        cout<<"YES"<<endl;
    return 0;
}