Description
The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They're planning to build some highways so that it will be possible to drive between any pair of towns without leaving the highway system. Flatopian towns are numbered from 1 to N. Each highway connects exactly two towns. All highways follow straight lines. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways. The Flatopian government wants to minimize the length of the longest highway to be built. However, they want to guarantee that every town is highway-reachable from every other town.Input
The first line of input is an integer T, which tells how many test cases followed. The first line of each case is an integer N (3 <= N <= 500), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 65536]) between village i and village j. There is an empty line after each test case.Output
For each test case, you should output a line contains an integer, which is the length of the longest road to be built such that all the villages are connected, and this value is minimum.Sample Input
1 3 0 990 692 990 0 179 692 179 0Sample Output
692解题思路：求最小生成树中的最大权是多少，我用的kruskal，用prime应该也可以做，但是为了熟悉kruskal 就》》#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 510;struct Edge{ int from; int to; int cost;}edge[MAXN * MAXN];int edgeNum;int pre[MAXN];int n;int find(int x){ while (x != pre[x]) { x = pre[x]; } return x;}int cmp(Edge ex, Edge ey){ return ex.cost < ey.cost;}void kruskal(){ sort(edge, edge + edgeNum, cmp); for (int i = 0; i < n; i++) { pre[i] = i; } int ret = 0; for (int i = 0; i < edgeNum; i++) { //Edge es = edge[i]; int a = find(edge[i].from); int b = find(edge[i].to); if (a != b) { ret = max(ret, edge[i].cost); pre[b] = a; } } printf("%d\n", ret);}int main(){ int t; scanf("%d", &t); while (t--) { edgeNum = 0; scanf("%d", &n); for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { scanf("%d", &edge[edgeNum].cost); edge[edgeNum].from = i; edge[edgeNum].to = j; edgeNum++; } } kruskal(); } return 0;}
