Solution Notes (Richard Peng): Two cows can reach each other if and only it is the case that, for each polygon, both cows are either in the interior of that polygon, or in the exterior. A community of cows is just a set of cows that are on the same "side" of each polygon. (A formal rigorous mathematical proof of this is nontrivial, but our geometric intuition suffices for this problem.)

If we can compute the polygons that a cow is in, we can sort the cows by the list of polygons that contain them, and pick the list which has most frequent occurence. As sorting the C cows takes O(ClogC) comparisons and each cow has N attributes, this step takes O(NlogN) time.

One way to check whether a cow is inside a polygon is to draw a ray from that cow in some direction that doesn't intersect a vertex of the polygon, and count the number of times this ray intersects the sides of the polygon. If the count is odd, the cow is contained, otherwise it's outside. This ray can be chosen arbitrarily, so there are a couple of approaches. One can always choose a horizontal ray for easy geometry, but then one has to be careful when vertices of the polygon lie on the ray. Alternatively, one can avoid the vertex issue entirely by choosing a line that will never contain any vertex (such as the ray from (x,y) through (x + 1, y + 1,000,000,000)) or by randomly choosing directions until one works. This can be done in time linear in the size of the polygon. As the total sizes of all polygons is N and there are C cows, this stage takes O(NC).

Here is Travis Hance's solution in C++:

#include <cstdio>
#include <map>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;

#define NFENCES_MAX 1005
#define NCOWS_MAX 1005

bool rayHits(long long cx, long long cy,
                         long long f1x, long long f1y,
                         long long f2x, long long f2y) {
    if ((f1y > cy) ^ (f2y > cy)) {
        return (f1y - f2y < 0) ^ (f2x * (f1y - cy) + f1x * (cy - f2y) > cx * (f1y - f2y));
    } else {
        return false;

map<pair<int,int>, vector<int> > pointMap;
pair<pair<int,int>, pair<int,int> > fences[NFENCES_MAX];
int cycle[NFENCES_MAX];

char parities[NCOWS_MAX][NFENCES_MAX];

char* parityptrs[NCOWS_MAX];
inline bool ptrcmp(char* a, char* b) {
    return strcmp(a, b) < 0;

int main() {

    int nFences, nCows;
    scanf("%d", &nFences);
    scanf("%d", &nCows);
    for (int i = 0; i < nFences; i++) {
        pair<int,int> p1, p2;
        scanf("%d", &p1.first);
        scanf("%d", &p1.second);
        scanf("%d", &p2.first);
        scanf("%d", &p2.second);
        fences[i] = pair<pair<int,int>, pair<int,int> >(p1, p2);

        cycle[i] = -1;
        memset(parities[i], 0, nCows);
    for (int i = 0; i < nCows; i++) {
        parityptrs[i] = parities[i];

    int nCycles = 0;
    for (int i = 0; i < nFences; i++) {
        if (cycle[i] == -1) {
            int j = i;
            pair<int,int> last = fences[i].first;
            do {
                cycle[j] = nCycles;
                last = fences[j].first == last ? fences[j].second : fences[j].first;
                vector<int> const& v = pointMap[last];
                j = (v[0] == j ? v[1] : v[0]);
            } while (j != i);

    for (int i = 0; i < nCows; i++) {
        int cowx, cowy;
        scanf("%d", &cowx);
        scanf("%d", &cowy);
        for (int j = 0; j < nFences; j++) {
            parities[i][cycle[j]] ^= (char)rayHits(cowx, cowy, fences[j].first.first, fences[j].first.second, fences[j].second.first, fences[j].second.second);
        for (int j = 0; j < nCycles; j++) {
            parities[i][j] = parities[i][j] ? '1' : '0';
        parities[i][nCycles] = '\0';

    sort(parityptrs, parityptrs + nCows, ptrcmp);

    int maxans = 0;
    int curcount = 0;
    for (int i = 0; i < nCows; i++) {
        if (i == 0 || strcmp(parityptrs[i], parityptrs[i-1]) != 0) {
            curcount = 1;
        } else {
        maxans = max(maxans, curcount);
    printf("%d\n", maxans);