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;
}