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-10;
const int N = 44;
const int M = 11;
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)) == 0 && sgn(dot(a - x, b - x)) <= 0;}
bool ptinpoly(Point x, Point *pt, int n) {
    pt[n] = pt[0];
    int wn = 0;
    for (int i = 0; i < n; i++) {
        if (onseg(x, pt[i], pt[i + 1])) return true;
        int dr = sgn(cross(pt[i + 1] - pt[i], x - pt[i]));
        int k1 = sgn(pt[i + 1].y - x.y);
        int k2 = sgn(pt[i].y - x.y);
        if (dr > 0 && k1 > 0 && k2 <= 0) wn++;
        if (dr < 0 && k2 > 0 && k1 <= 0) wn--;
    }
    return wn != 0;
}

struct MFS {
    int fa[N], cnt;
    void init() { for (int i = 0; i < N; i++) fa[i] = i; cnt = 0;}
    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[111];
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 < 0 && t1 * t2 < 0;
}

bool polyint(int a, int b) {
    poly[a][sz[a]] = poly[a][0];
    poly[b][sz[b]] = poly[b][0];
    for (int i = 0; i < sz[a]; i++) {
        for (int j = 0; j < sz[b]; j++) {
            if (ssint(poly[a][i], poly[a][i + 1], poly[b][j], poly[b][j + 1])) return true;
        }
    }
    return false;
}

bool test(int a, int b) {
    for (int i = 0; i < sz[a]; i++) if (ptinpoly(poly[a][i], poly[b], sz[b])) return true;
    for (int i = 0; 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 = 0; i < n; i++) {
            gets(buf);
            p = strtok(buf, " ");
            for (sz[i] = 0; 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 = 0; j < i; j++) if (test(i, j)) {
                mfs.merge(i, j);
                //cout << i << ' ' << j << endl;
            }
        }
        cout << n - mfs.cnt << endl;
    }
    return 0;
}

——written by Lyon