HDU 6982(Road Discount-wqs二分)

给你一个无向图$n$个点$m$条边，每个边有一个代价$c_i$以及折扣价$d_i$，现在要求对每个$k(0\le k \le n-1)$，求恰取$k$条折扣价的最小生成树的代价是多少。
$n\le 1e3,m\le2e5,c_i,d_i\le 1e3$

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) {\
            For(j,m-1) cout<<a[i][j]<<' ';\
            cout<<a[i][m]<<endl;\
            }
#pragma comment(linker,"/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
#define sqr(x) (x*x)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
  int x=0,f=1; char ch=getchar();
  while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
  while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
  return x*f;
} 

#define MAXN (400000+10)
int n, m, u[MAXN],v[MAXN],id[MAXN],w[MAXN],d[MAXN],c[MAXN];
bool b[MAXN<<1];
int iid[MAXN],t0=0;
class bingchaji
{
public:
  #define MAXN (1010)
  int father[MAXN],n,cnt;
  void mem(int _n)
  {
    n=cnt=_n;
    For(i,n) father[i]=i;
  }
  int getfather(int x) 
  {
    if (father[x]==x) return x;
    
    return father[x]=getfather(father[x]);
  }
  void unite(int x,int y)
  {
    x=getfather(x);
    y=getfather(y);
    if (x^y) {
      --cnt;
    
      father[x]=y;
    }
  }
  bool same(int x,int y)
  {
    return getfather(x)==getfather(y);
  }
}S;
bool cmp(int i,int j) {
  return w[i]<w[j]||w[i]==w[j]&&i<j;
}
int an[1010];
int M;
int work(int m,bool flag,int &p1) {
  if(!flag){
    For(i,m) id[i]=i;
    sort(id+1,id+1+m,cmp);
  }
  S.mem(n);
  int ans=0;
  For(p,m) {
    int i=id[p];
    if(!S.same(u[i],v[i])){
      S.unite(u[i],v[i]);
      
      p1+=i<=M;
      if(!flag)b[i]=1;
      ans+=w[i];
    }
  }
  return ans;
}
int main()
{
//  freopen("j.in","r",stdin);
//  freopen(".out","w",stdout);
  int T=read();
  For(kcase,T) {
    n=read(),m=read();
    For(i,m) u[i]=read(),v[i]=read(),c[i]=read(),d[i]=read(),b[i]=0;
    int p1=0;
    Rep(i,n) an[i]=INF;
    M=m;
    t0=0;
    
    For(i,m) w[i]=c[i];
    an[n-1]=work(m,0,p1); // no discount
    For(i,m) w[i]=d[i];
    an[0]=work(m,0,p1);
    M=0;
    For(i,m) if(b[i]) {
      ++M;
      u[M]=u[i],v[M]=v[i];
      d[M]=d[i],c[M]=c[i];
    }
    m=M;
    For(i,M) u[i+M]=u[i],v[i+M]=v[i],w[i]=c[i],w[i+M]=d[i];
    For(i,M) iid[i]=i; sort(iid+1,iid+1+M,cmp);
    For(i,M) iid[i+M]=i+M; sort(iid+M+1,iid+1+2*M,cmp);
    
    int pr=0;
    Rep(mid,1002){
      For(i,m) w[i]=c[i],w[i+m]=d[i]+mid;
      
      int p1=1,p2=1;
      For(ttt,2*M) {
        if(p1>M) id[ttt]=iid[p2+M],++p2;
        else if(p2>M) id[ttt]=iid[p1],++p1;
        else {
          if(w[iid[p1]]>w[iid[p2+M]]) id[ttt]=iid[p2+M],++p2;
          else id[ttt]=iid[p1],++p1;
        }
      }
      
      
      p1=0;
      int ans=work(2*M,1,p1);
      ans-mid*(n-1-p1);
      
      gmin(an[p1],ans-mid*(n-1-p1))
      if(pr+1<p1) {
        Fork(j,pr+1,p1-1) gmin(an[j],ans-mid*(n-1-j))
      }
      pr=p1;
    }
    RepD(i,n-2) {
      gmin(an[i],an[i+1])
    }
    RepD(i,n-1) printf("%d\n",an[i]);
  }
  return 0;
}