Cycling Roads URAL1966 并查集
1966. Cycling Roads
Time limit: 1.0 second
Memory limit: 64 MB
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
Tags: none
Difficulty: 587 Printable version Submit solution Discussion (2)
My submissions All submissions (894) All accepted submissions (175) Solutions rating (109)
My submissions All submissions (894) All accepted submissions (175) Solutions rating (109)
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; }
posted on 2013-10-13 18:46 keep trying 阅读(245) 评论(1) 编辑 收藏 举报