(kruscal12.3.5)POJ 2485 Highways(使用kruscal来计算最小生成树的最大边)

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct edge {int begin;int end;int weight;};const int maxn = 1100;int father[maxn];edge e[maxn * maxn];int map[maxn][maxn];int n;int find(int x) {if (x == father[x]) {return x;}father[x] = find(father[x]);return father[x];}int kruscal(int count) { //使用kruscal算法来生成最小生成树并计算带权路径和int i;int sum = 0; //用sum来记录最小s生成树的边权和int done = 0;for (i = 1; i < maxn; ++i) {father[i] = i;}for (i = 1; i <= count; ++i) { //枚举有序边集中的每一条边int fx = find(e[i].begin);int fy = find(e[i].end);if (fx != fy) { //若第k条边的两个端点i,j 分别属于两颗不同的子树father[fx] = fy; //则将节点i所在的子树并入节点j所在的子树中sum += e[i].weight;done++;//计算目前树中的边数if (done == n - 1) {break;}}}printf("%d\n",e[i].weight);//输出最小生成树的最大边return sum;}bool compare(const edge& a, const edge& b) {return a.weight < b.weight;}//以上是用kruscal算法来解决问题的基本模板.....int main() {int t;scanf("%d",&t);while(t--){scanf("%d",&n);int i,j;for(i = 1 ; i <= n ; ++i){for(j = 1 ; j <= n ; ++j){scanf("%d",&map[i][j]);}}int count = 0;for(i = 1 ; i <= n ; ++i){//*******下标要是从0开始会WA。。(好吧,我也不知道是为什么)for(j = i+1 ; j <= n ; ++j){e[count].begin = i;e[count].end = j;e[count++].weight = map[i][j];}}sort(e+1,e+count+1,compare);kruscal(count);}return 0;}