POJ 2253 Difference of clustering

Test instructions: Gives a pile of points, the minimum value of the maximum distance from the point of origin to the end of all paths. (The meaning is oneself Baidu bar ...) ）

Solution: Use the maximum value of the adjacent point as the weight value instead of the distance of the path to run the shortest or smallest spanning tree. And then I wrote a Dijkstra that I thought was an optimized one, but it was like prim--ah, pretty much.

In short, with priority queue maintenance weights for wide search ... And then pay g++ always wa also don't know why ... C + + has been handed over ...

Code:

#include <stdio.h> #include <iostream> #include <algorithm> #include <string> #include < string.h> #include <math.h> #include <limits.h> #include <time.h> #include <stdlib.h># include<map> #include <queue> #include <set> #include <stack> #include <vector> #define LL Long longusing namespace std;struct node{int x, y;}    p[205];struct node1{int point;    int step;    Node1 (int, int step): Point, Step (step) {} node1 () {} BOOL operator < (const node1 &AMP;TMP) const    {return step > tmp.step; }};int N;int Caldis (Node A, Node B) {return (a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (A.Y-B.Y);}    int dis[205][205];vector <int> edge[205];bool vis[205];d ouble bfs () {memset (Vis, 0, sizeof vis);    Priority_queue <node1> Q;    Q.push (Node1 (0, 0.0));        while (!q.empty ()) {Node1 tmp = Q.top ();        Q.pop ();        Vis[tmp.point] = 1; if (Tmp.point = = 1) return Tmp.step; for (int i = 0; i < edge[tmp.point].size (); i++) {if (!vis[edge[tmp.point][i])) Q.push (Node1 (edge[tmp).        Point][i], Max (Tmp.step, Dis[tmp.point][edge[tmp.point][i]));    }}}int Main () {int CSE = 1; while (~SCANF ("%d", &n) && N) {for (int i = 0; i < n; i++) scanf ("%d%d", &p[i].x, &        AMP;P[I].Y);        int maxn = 0.0;        for (int i = 0; i < n; i++) edge[i].clear ();                for (int i = 0, i < n; i++) for (int j = 0; J < N; j + +) {if (i = = j) Continue;                int tmp = Caldis (P[i], p[j]);                Edge[i].push_back (j);            DIS[I][J] = tmp;        } Double ans = BFS ();    printf ("Scenario #%d\nfrog Distance =%.3lf\n\n", cse++, sqrt (ans)); } return 0;}

　　

POJ 2253 Difference of clustering