一開始還覺得很難，結果。一發過了。

其實觀察可以發現就是一個最小生成樹。

對於前k條邊，我們去連最小生成樹，顯然這時比較的是自己想的花費。

然後剩下的就是老師講的花費比較，再去連最小生成樹。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 1e4+5;
const int M = 2e4+5;
const LL Mod = 1e9+7;
#define rg register
#define
 
 pi acos(-1)
#define INF 1e9
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '
 
-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace
 
 FASTIO;
void FRE(){/*freopen("data1.in","r",stdin);
freopen("data1.out","w",stdout);*/}

struct Node{int st,to,dis1,dis2;}e[M];
bool cmp1(Node a,Node b){return a.dis1 < b.dis1;}
bool cmp2(Node a,Node b){return a.dis2 < b.dis2;}
int fa[N];
int Find(int x)
{
    return x == fa[x] ? x : fa[x] = Find(fa[x]);
}
int main()
{
    int n,k,m;n = read(),k = read(),m = read();
    for(int i = 1;i <= n;++i) fa[i] = i;
    for(int i = 1;i <= m;++i) e[i].st = read(),e[i].to = read(),e[i].dis1 = read(),e[i].dis2 = read();
    sort(e+1,e+m+1,cmp1);
    int num = k,mx = -1;
    for(int i = 1;i <= m && num > 0;++i)
    {
        int x = e[i].st,y = e[i].to,dis = e[i].dis1;
        int xx = Find(x),yy = Find(y);
        if(xx != yy)
        {
            fa[xx] = yy;
            mx = max(mx,dis);
            num--;
        }
    }
    sort(e+1,e+m+1,cmp2);
    for(int i = 1;i <= m;++i)
    {
        int x = e[i].st,y = e[i].to,dis = e[i].dis2;
        int xx = Find(x),yy = Find(y);
        if(xx != yy)
        {
            fa[xx] = yy;
            mx = max(mx,dis);
        }
    }
    printf("%d\n",mx);
    system("pause");    
}