poj 2536 Gopher II 最大二分匹配

       这题难点建图。

       建图方式如下：要是在给定的s时间内能跑到洞穴（判断两点距离），那么那两个点是连通的，否则就不连通

 

#include<iostream>

using namespace std;



const int MAX = 300;



struct coordinate

{

	double x,y;

};



coordinate gopher[MAX];

coordinate holes[MAX];



bool arcs[MAX][MAX];

bool isvisit[MAX];

int match[MAX];

int n, m, s, v;



bool find(int u)

{

	for (int i = 1; i <= m; i++)

		if (arcs[u][i] && !isvisit[i])

		{

			isvisit[i] = true;

			if (!match[i] || find(match[i]))

			{

				match[i] = u;

				return true;

			}

		}



	return false;

}



int main()

{

	while (cin >> n >> m >> s >> v)

	{

		double d = s*v*s*v;

		for (int i = 0; i < n; i++)

			cin >> gopher[i].x >> gopher[i].y;



		for (int i = 0; i < m; i++)

			cin >> holes[i].x >> holes[i].y;



		memset(arcs, false, sizeof(arcs));

		memset(match, 0, sizeof(match));

		for (int i = 0; i < n; i++)

			for (int j = 0; j < m; j++)

				if ((holes[j].x - gopher[i].x)*(holes[j].x - gopher[i].x) + (holes[j].y - gopher[i].y)*(holes[j].y - gopher[i].y) <= d)

					arcs[i+1][j+1] = true;



		int ans = 0;

		for (int i = 1; i <= n; i++)

		{

			memset(isvisit, false, sizeof(isvisit));

			if (find(i))

				ans++;

		}



		cout << n - ans << endl;

	}

	return 0;

}