# 2021“MINIEYE杯”中国大学生算法设计超级联赛（1）1009. KD-Graph（并查集）

Problem Description

Let’s call a weighted connected undirected graph of n vertices and m edges KD-Graph, if the
following conditions fulfill:

* n vertices are strictly divided into K groups, each group contains at least one vertice

* if vertices p and q ( p ≠ q ) are in the same group, there must be at least one path between p and q meet the max value in this path is less than or equal to D.

* if vertices p and q ( p ≠ q ) are in different groups, there can’t be any path between p and q meet the max value in this path is less than or equal to D.

You are given a weighted connected undirected graph G of n vertices and m edges and an integer K.

Your task is find the minimum non-negative D which can make there is a way to divide the n vertices into K groups makes G satisfy the definition of KD-Graph.Or ?1if there is no such D exist.

Input

The first line contains an integer T (1≤ T ≤5) representing the number of test cases.
For each test case , there are three integers n,m,k(2≤n≤100000,1≤m≤500000,1≤k≤n) in the first line.
Each of the next m lines contains three integers u,v and c (1≤v,u≤n,v≠u,1≤c≤109) meaning that there is an edge between vertices u and v with weight c.

Output

For each test case print a single integer in a new line.

Sample Input

``````2
3 2 2
1 2 3
2 3 5
3 2 2
1 2 3
2 3 3
``````

Sample Output

``````3
-1
``````

``````#include <bits/stdc++.h>
#define N 100005
#define M 5000005
using namespace std;
int n, m, k;
struct edge{
int x, y, z;
} e[M];
bool cmp(edge a, edge b) {
return a.z < b.z;
}
int fa;
int get(int x) {
if(x == fa[x]) return x;
return fa[x] = get(fa[x]);
}
int main() {
//freopen("1.in", "r", stdin);
int t;
cin >> t;
while(t--) {
scanf("%d%d%d", &n, &m, &k);
for(int i = 1; i <= m; i++) {
scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].z);
}
for(int i = 1; i <= n; i++) fa[i] = i;
sort(e + 1, e + m + 1, cmp);
int conn = n;
int D = 0x3f3f3f3f;
bool flag = 1;
if(k == n) {
D = 0;
}
for(int i = 1; i <= m; i++) {
if(D == 0) break;
int x = e[i].x, y = e[i].y, z = e[i].z;
int fx = get(x), fy = get(y);
//cout << z << endl;
if(fx == fy) {
continue;
} else {
if(conn > k + 1) {
//不用管
} else if(conn == k + 1) {
D = z;
} else {
if(z <= D) {
flag = 0;
}
break;
}
fa[fx] = fy;
conn--;
}
}
if(!flag || D == 0x3f3f3f3f) cout << -1 << endl;
else cout << D << endl;
}
return 0;
}
``````

2021“MINIEYE杯”中国大学生算法设计超级联赛（1）1009. KD-Graph（并查集）

(0)
(0)