HDU 2389 二分图最大匹配之Hopcroft-Karp优化 O(n^0.5*m)

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3010;
const int MAXM = 3010*3010;
const int INF = 0x3f3f3f3f;
struct Edge {
	int v;
	int next;
} edge[MAXM];
struct node {
	double x, y;
	double v;
} a[MAXN], b[MAXN];
int nx, ny;
int cnt;
int t;
int dis;
int first[MAXN];
int xlink[MAXN], ylink[MAXN]; /*xlink[i]表示左集合顶点所匹配的右集合顶点序号，ylink[i]表示右集合i顶点匹配到的左集合顶点序号。*/
int dx[MAXN], dy[MAXN]; /*dx[i]表示左集合i顶点的距离编号，dy[i]表示右集合i顶点的距离编号*/
int vis[MAXN]; //寻找增广路的标记数组
void init() {
	cnt = 0;
	memset(first, -1, sizeof(first));
	memset(xlink, -1, sizeof(xlink));
	memset(ylink, -1, sizeof(ylink));
}
void read_graph(int u, int v) {
	edge[cnt].v = v;
	edge[cnt].next = first[u], first[u] = cnt++;
}
int bfs() {
	queue<int> q;
	dis = INF;
	memset(dx, -1, sizeof(dx));
	memset(dy, -1, sizeof(dy));
	for(int i = 0; i < nx; i++)	{
		if(xlink[i] == -1)		{
			q.push(i);
			dx[i] = 0;
		}
	}
	while(!q.empty())	{
		int u = q.front();
		q.pop();
		if(dx[u] > dis) break;
		for(int e = first[u]; e != -1; e = edge[e].next)		{
			int v = edge[e].v;
			if(dy[v] == -1)			{
				dy[v] = dx[u] + 1;
				if(ylink[v] == -1) dis = dy[v];
				else				{
					dx[ylink[v]] = dy[v]+1;
					q.push(ylink[v]);
				}
			}
		}
	}
	return dis != INF;
}
int find(int u) {
	for(int e = first[u]; e != -1; e = edge[e].next){
		int v = edge[e].v;
		if(!vis[v] && dy[v] == dx[u]+1)	{
			vis[v] = 1;
			if(ylink[v] != -1 && dy[v] == dis) continue;
			if(ylink[v] == -1 || find(ylink[v])){
				xlink[u] = v, ylink[v] = u;
				return 1;
			}
		}
	}
	return 0;
}
int MaxMatch() {
	int ans = 0;
	while(bfs())	{
		memset(vis, 0, sizeof(vis));
		for(int i = 0; i < nx; i++) if(xlink[i] == -1)	{
				ans += find(i);
			}
	}
	return ans;
}
double dist(const node a, const node b) {
	return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}
void read_case() {
	init();
	int Time;
	scanf("%d", &Time);
	scanf("%d", &nx);
	for(int i = 0; i < nx; i++)	{
		scanf("%lf%lf%lf", &a[i].x, &a[i].y, &a[i].v);
	}
	scanf("%d", &ny);
	for(int i = 0; i < ny; i++)	{
		scanf("%lf%lf", &b[i].x, &b[i].y);
	}
	for(int i = 0; i < nx; i++)	{
		for(int j = 0; j < ny; j++)		{
			double limit = a[i].v*Time;
			double s = dist(a[i], b[j]);
			if(s <= limit) read_graph(i, j);
		}
	}
}
void solve() {
	read_case();
	int ans = MaxMatch();
	printf("%d\n\n", ans);
}
int main() {
	int T, times = 0;
	scanf("%d", &T);
	while(T--)	{
		printf("Scenario #%d:\n", ++times);
		solve();
	}
	return 0;
}