(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;
}