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URAL - 1966 - Cycling Roads(并检查集合 + 判刑线相交)

时间:2015-09-30 21:02:48      阅读:198      评论:0      收藏:0      [点我收藏+]

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意甲冠军:n 积分,m 边缘(1 ≤ m < n ≤ 200),问:是否所有的点连接(两个边相交。该 4 点连接)。

主题链接:http://acm.timus.ru/problem.aspx?space=1&num=1966

——>>对于每条边,边上的两端点并入集合,枚举边与边。推断他们是否相交,是的话各点并入集合,最后看集合内元素的个数是否为n。。

#include <cstdio>
#include <cmath>

const int MAXN = 200 + 10;
const double EPS = 1e-8;

struct POINT
{
    double x;
    double y;

    POINT(){}

    POINT(double x, double y) : x(x), y(y){}
} p[MAXN];

int n, m;
int fa[MAXN], cnt[MAXN];
int u[MAXN], v[MAXN];

typedef POINT Vector;
Vector operator - (Vector A, Vector B)
{
    return Vector(A.x - B.x, A.y - B.y);
}

int Dcmp(double x)
{
    if (fabs(x) < EPS) return 0;
    return x > 0 ? 1 : -1;
}

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 SegmentProperIntersection(POINT a1, POINT a2, POINT b1, POINT b2)
{
    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;
}

bool OnSegment(POINT p, POINT a1, POINT a2)
{
    return Dcmp(Cross(a1 - p, a2 - p)) == 0 && Dcmp(Dot(a1 - p, a2 - p)) < 0;
}

void Init()
{
    for (int i = 1; i <= n; ++i)
    {
        fa[i] = i;
        cnt[i] = 1;
    }
}

int Find(int x)
{
    return x == fa[x] ? x : (fa[x] = Find(fa[x]));
}

void Union(int x, int y)
{
    int xroot = Find(x);
    int yroot = Find(y);

    if (xroot != yroot)
    {
        fa[yroot] = xroot;
        cnt[xroot] += cnt[yroot];
    }
}

void Read()
{
    for (int i = 1; i <= n; ++i)
    {
        scanf("%lf%lf", &p[i].x, &p[i].y);
    }
    for (int i = 0; i < m; ++i)
    {
        scanf("%d%d", u + i, v + i);
        Union(u[i], v[i]);
        for (int j = 1; j <= n; ++j)
        {
            if (j != u[i] && j != v[i] && OnSegment(p[j], p[u[i]], p[v[i]]))
            {
                Union(j, u[i]);
            }
        }
    }
}

void Merge()
{
    for (int i = 0; i < m; ++i)
    {
        for (int j = i + 1; j < m; ++j)
        {
            if (SegmentProperIntersection(p[u[i]], p[v[i]], p[u[j]], p[v[j]]))
            {
                Union(u[i], u[j]);
            }
        }
    }
}

void Output()
{
    cnt[Find(1)] == n ?

puts("YES") : puts("NO"); } int main() { while (scanf("%d%d", &n, &m) == 2) { Init(); Read(); Merge(); Output(); } return 0; }



版权声明:本文博主原创文章,博客,未经同意不得转载。

URAL - 1966 - Cycling Roads(并检查集合 + 判刑线相交)

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原文地址:http://www.cnblogs.com/lcchuguo/p/4850194.html

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