uva 11665 Chinese Ink (几何+并查集)

UVA 11665

　　随便给12的找了一道我没做过的几何基础题。这题挺简单的，不过uva上通过率挺低，通过人数也不多。

　　题意是要求给出的若干多边形组成多少个联通块。做的时候要注意这题是不能用double浮点类型的，然后判多边形交只需要两个条件，存在边规范相交，或者存在一个多边形上的顶点在另一个多边形上或者在多边形内。

代码如下：

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 const double EPS = 1e-;
 const int N = ;
 const int M = ;
 inline int sgn(double x) { return (x > EPS) - (x < -EPS);}

 struct Point {
     int x, y;
     Point() {}
     Point(int x, int y) : x(x), y(y) {}
     Point operator + (Point a) { return Point(x + a.x, y + a.y);}
     Point operator - (Point a) { return Point(x - a.x, y - a.y);}
     Point operator * (int p) { return Point(x * p, y * p);}
     Point operator / (int p) { return Point(x / p, y / p);}
 } ;

 inline int cross(Point a, Point b) { return a.x * b.y - a.y * b.x;}
 inline int dot(Point a, Point b) { return a.x * b.x + a.y * b.y;}

 inline bool onseg(Point x, Point a, Point b) { return sgn(cross(a - x, b - x)) ==  && sgn(dot(a - x, b - x)) <= ;}
 bool ptinpoly(Point x, Point *pt, int n) {
     pt[n] = pt[];
     int wn = ;
     for (int i = ; i < n; i++) {
         if (onseg(x, pt[i], pt[i + ])) return true;
         int dr = sgn(cross(pt[i + ] - pt[i], x - pt[i]));
         int k1 = sgn(pt[i + ].y - x.y);
         int k2 = sgn(pt[i].y - x.y);
         if (dr >  && k1 >  && k2 <= ) wn++;
         if (dr <  && k2 >  && k1 <= ) wn--;
     }
     return wn != ;
 }

 struct MFS {
     int fa[N], cnt;
     void init() { for (int i = ; i < N; i++) fa[i] = i; cnt = ;}
     int find(int x) { return fa[x] = fa[x] == x ? x : find(fa[x]);}
     void merge(int x, int y) {
         int fx = find(x);
         int fy = find(y);
         if (fx == fy) return ;
         cnt++;
         fa[fx] = fy;
     }
 } mfs;

 Point poly[N][M];
 char buf[];
 int sz[N];

 bool ssint(Point a, Point b, Point c, Point d) {
     int s1 = sgn(cross(a - c, b - c));
     int s2 = sgn(cross(a - d, b - d));
     int t1 = sgn(cross(c - a, d - a));
     int t2 = sgn(cross(c - b, d - b));
     return s1 * s2 <  && t1 * t2 < ;
 }

 bool polyint(int a, int b) {
     poly[a][sz[a]] = poly[a][];
     poly[b][sz[b]] = poly[b][];
     for (int i = ; i < sz[a]; i++) {
         for (int j = ; j < sz[b]; j++) {
             if (ssint(poly[a][i], poly[a][i + ], poly[b][j], poly[b][j + ])) return true;
         }
     }
     return false;
 }

 bool test(int a, int b) {
     for (int i = ; i < sz[a]; i++) if (ptinpoly(poly[a][i], poly[b], sz[b])) return true;
     for (int i = ; i < sz[b]; i++) if (ptinpoly(poly[b][i], poly[a], sz[a])) return true;
     if (polyint(a, b)) return true;
     return false;
 }

 int main() {
     //freopen("in", "r", stdin);
     int n;
     while (cin >> n && n) {
         gets(buf);
         char *p;
         mfs.init();
         for (int i = ; i < n; i++) {
             gets(buf);
             p = strtok(buf, " ");
             for (sz[i] = ; p; sz[i]++) {
                 sscanf(p, "%d", &poly[i][sz[i]].x);
                 p = strtok(NULL, " ");
                 sscanf(p, "%d", &poly[i][sz[i]].y);
                 p = strtok(NULL, " ");
             }
             //cout << sz[i] << endl;
             for (int j = ; j < i; j++) if (test(i, j)) {
                 mfs.merge(i, j);
                 //cout << i << ' ' << j << endl;
             }
         }
         cout << n - mfs.cnt << endl;
     }
     return ;
 }

——written by Lyon