Ultraviolet A 534-Frogger (Kruskal extension)

Ultraviolet A 534-Frogger

Question Link

Given some points, a path is required to jump from the first point to the second point, and the maximum distance on the path is the smallest.

Idea: Use the Kruskal algorithm to add the edge of the smallest weight value each time to determine whether the edge can be connected to two points. If yes, the current weight is that the answer complexity is O (n ^ 2log (n ))

But in fact, I can use Floyd to create O (N ^ 3 .. However, the efficiency is not as good as the above method.

Code:

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int N = 205;struct Point {int x, y;void read() {scanf("%d%d", &x, &y);}} p[N];double dis(Point a, Point b) {int dx = a.x - b.x;int dy = a.y - b.y;return sqrt(dx * dx + dy * dy);}struct Edge {int u, v;double d;Edge() {}Edge(int u, int v) {this->u = u;this->v = v;d = dis(p[u], p[v]);}bool operator < (const Edge& c) const {return d < c.d;}} E[N * N];int n, en, parent[N];int find(int x) {return x == parent[x] ? x : parent[x] = find(parent[x]);}int main() {int cas = 0;while (~scanf("%d", &n) && n) {en = 0;for (int i = 0; i < n; i++) {parent[i] = i;p[i].read();for (int j = 0; j < i; j++)E[en++] = Edge(i, j);}sort(E, E + en);for (int i = 0; i < en; i++) {int pa = find(E[i].u);int pb = find(E[i].v);if (pa != pb)parent[pa] = pb;if (find(0) == find(1)) {printf("Scenario #%d\nFrog Distance = %.3lf\n\n", ++cas, E[i].d);break;}}}return 0;}

Ultraviolet A 534-Frogger (Kruskal extension)