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Ural 1966 Cycling Roads
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Cycling Roads
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Description
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.
Sample Input
| input | output |
|---|---|
4 20 01 01 10 11 34 2 | YES |
4 30 01 01 10 11 22 13 4 | NO |
3 20 01 01 11 33 2 | YES |
这道题主要是判相交，只要相交就把它压入并查集，一开始我是用了cnt去记录已经相交的节点，后来发现不行，因为新加如的一条线如果加进去了，它的另外一个端点也会加入，导致cnt记录的数值不准。于是用了另外一个数组c[i]去记录以i为根的所有子节点的个数。
在判断相交这里，一开始没有注意到新加入一条线段时，应该判断所有点是否在该线段上，如果端点在该线段上，则把它加入，加了OnSegment()判断之后就AC了。
#include<cstdio>#include<cmath>#include<iostream>using namespace std;#define maxn 205struct point{ double x,y; point(double x = 0,double y = 0):x(x),y(y){}}p[maxn];struct Line{ point a,b; int pos1,pos2; Line(){} Line(point x,point y,int ppos1,int ppos2){ a = x; b = y; pos1 = ppos1; pos2 = ppos2;}}line[maxn];int n,m,cnt;int par[maxn];int c[maxn];typedef point Vector;Vector operator +(Vector A,Vector B){ return Vector(A.x+B.x, A.y+B.y); }Vector operator -(Vector A,Vector B) { return Vector(A.x-B.x,A.y-B.y); }Vector operator *(Vector A,double p) { return Vector(A.x*p,A.y*p); }Vector operator /(Vector A,double p){ return Vector(A.x/p,A.y/p); }const double eps = 1e-10;int dcmp(double x){ if(fabs(x) < eps) return 0; else return x < 0? -1:1;}bool operator == (const point &a,const point &b){ return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;}double dot(Vector A,Vector B){ return A.x*B.x + A.y*B.y; }double cross(Vector A,Vector B){ return A.x*B.y-A.y*B.x; }bool OnSegment(point p,Line l){ return dcmp(cross(l.a-p,l.b-p)) == 0 && dcmp(dot(l.a-p,l.b-p)) < 0;}bool SegmentProperIntersection(Line l1,Line l2){ point a1 = l1.a; point a2 = l1.b; point b1 = l2.a; point b2 = l2.b; double c1 = cross(a2-a1,b1-a1); double c2 = cross(a2-a1,b2-a1); double c3 = cross(b2-b1,a1-b1); double c4 = cross(b2-b1,a2-b1); return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;}void init(){ for(int i = 1; i <= n; i++) c[i] = 1; for(int i = 0; i < maxn;i++) par[i] = i;}int Find(int x){ if(par[x] != x) { return par[x]=Find(par[x]); } else return x;}void Merge(int a,int b){ int t1 = Find(a); int t2 = Find(b); if(t1 != t2) { par[t2] = t1; c[t1] += c[t2]; //printf("%d %d merge\n",a,b); //return 1; } //return 0;}void input(){ int x,y; for(int i = 1; i <= n; i++) { double x,y; scanf("%lf%lf",&x,&y); p[i] = point(x,y); } for(int i = 0; i < m; i++) { scanf("%d%d",&x,&y); line[i] = Line(p[x],p[y],x,y); for(int j = 1; j <= n; j++) { if(OnSegment(p[j],line[i])) Merge(j,x); } Merge(x,y); }}void deal(){ for(int i = 0; i < m; i++) { for(int j = i + 1; j < m; j++) { if(SegmentProperIntersection(line[i],line[j])) { Merge(line[j].pos1,line[i].pos1); //Merge(line[j].pos2,line[i].pos1); } } }}int main(){ //freopen("input.txt","r",stdin); while(scanf("%d%d",&n,&m) == 2) { init(); input(); deal(); if(c[Find(1)] == n) printf("YES\n"); else printf("NO\n"); } return 0;}
Ural 1966 Cycling Roads
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