UVA 1411 - Ants(二分图完美匹配)
2014-11-06 10:00
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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 , b ; 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; }
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