容易想到,对于 当前这个图,如果要得到恰好need条边,且边权和最小,
那么就要控制白边在生成图中的数量…那么想到如果给所有白边加一个值,
跑克鲁斯卡尔的话白边数量就会变化,那么考虑二分即可。
c++代码如下:
#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i = x ; i <= y ;++ i)
#define repd(i,x,y) for(register int i = x ; i >= y; -- i)
using namespace std;
typedef long long ll;
template<typename T>inline void read(T&x)
{
char c;int sign = 1;x = 0;
do { c = getchar(); if(c == '-') sign = -1; }while(!isdigit(c));
do { x = x * 10 + c - '0'; c = getchar(); }while(isdigit(c));
x *= sign;
}
const int N = 1e5+50, inf = 1e9+7;
int n,m,need,ans,now,p[N];
struct Edge { int u,v,w,op; }e[N];
const bool cmp(Edge a,Edge b) { return a.w < b.w || a.w == b.w && a.op < b.op; }
int get_fa(int x) { return p[x] == x ? x : p[x] = get_fa(p[x]); }
inline bool check(int w)
{
rep(i,1,n) p[i] = i; now = 0;int c = 0;
rep(i,1,m) if(e[i].op == 0) e[i].w += w;
sort(e + 1,e + 1 + m,cmp);
rep(i,1,m)
{
int x = get_fa(e[i].u),y = get_fa(e[i].v);
if(x != y)
{
if(e[i].op == 0 ) ++ c ;
p[x] = y; now += e[i].w;
}
}
rep(i,1,m) if(e[i].op == 0) e[i].w -= w;
return c >= need;
}
int main()
{
read(n); read(m); read(need);
rep(i,1,m) read(e[i].u), read(e[i].v), read(e[i].w), read(e[i].op);
rep(i,1,m) e[i].u ++ , e[i].v ++ ;
int l = -105,r = 105,mid;
while( l <= r )
{
if(check(mid = l + r >> 1)) ans = now - need*mid,l = mid + 1;
else r = mid - 1;
}
cout << ans << endl;
return 0;
}