UVA 1411 - Ants

题目链接

题意：给定一些黑点白点，要求一个黑点连接一个白点，而且全部线段都不相交

思路：二分图完美匹配，权值存负的欧几里得距离，这种话，相交肯定比不相交权值小，所以做一次完美匹配就能够了

代码：

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

const int MAXNODE = 105;

typedef double Type;

const Type INF = 0x3f3f3f3f;

struct KM {

	int n;

	Type g[MAXNODE][MAXNODE];

	Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];

	int left[MAXNODE];

	bool S[MAXNODE], T[MAXNODE];

	void init(int n) {

		this->n = n;

	}

	void add_Edge(int u, int v, Type val) {

		g[u][v] = val;

	}

	bool dfs(int i) {

		S[i] = true;

		for (int j = 0; j < n; j++) {

			if (T[j]) continue;

			double tmp = Lx[i] + Ly[j] - g[i][j];

			if (fabs(tmp) < 1e-9) {

				T[j] = true;

				if (left[j] == -1 || dfs(left[j])) {

					left[j] = i;

					return true;

				}

			} else slack[j] = min(slack[j], tmp);

		}

		return false;

	}

	void update() {

		double a = INF;

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

			if (!T[i]) a = min(a, slack[i]);

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

			if (S[i]) Lx[i] -= a;

			if (T[i]) Ly[i] += a;

		}

	}

	void km() {

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

			left[i] = -1;

			Lx[i] = -INF; Ly[i] = 0;

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

				Lx[i] = max(Lx[i], g[i][j]);

		}

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

			for (int j = 0; j < n; j++) slack[j] = INF;

			while (1) {

				for (int j = 0; j < n; j++) S[j] = T[j] = false;

				if (dfs(i)) break;

				else update();

			}

		}

	}

} gao;

const int N = 105;

int n;

struct Point {

	double x, y;

	void read() {

		scanf("%lf%lf", &x, &y);

	}

} a[N], b[N];

double dis(Point a, Point b) {

	double dx = a.x - b.x;

	double dy = a.y - b.y;

	return sqrt(dx * dx + dy * dy);

}

int main() {

	int bo = 0;

	while (~scanf("%d", &n)) {

		if (bo) printf("\n");

		else bo = 1;

		gao.init(n);

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

			a[i].read();

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

			b[i].read();

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

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

				gao.add_Edge(i, j, -dis(b[i], a[j]));

		gao.km();

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

			printf("%d\n", gao.left[i] + 1);

	}

	return 0;

}