bzoj 1196

http://www.lydsy.com/JudgeOnline/problem.php?id=1196

二分+并查集

一共有2*M条路径，我们首先将这2*M条路径按费用排序。

然后二分最大费用的公路mid

变成判断性问题：能否只用第1到第mid条公路，使得生成树至少包含K条一级公路。

因为这时候已经跟费用无关了，我们优先选取一级公路即可。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj

using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define fill(a,l,r,v) fill(a+l,a+r+1,v)
#define re(i,a,b)  for(i=(a);i<=(b);i++)
#define red(i,a,b) for(i=(a);i>=(b);i--)
#define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

const DB EPS=1e-;
inline int sgn(DB x){if(abs(x)<EPS)return ;return(x>)?:-;}
const DB Pi=acos(-1.0);

inline int gint()
  {
        int res=;bool neg=;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return ;
        if(z=='-'){neg=;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());
        return (neg)?-res:res;
    }
inline LL gll()
  {
      LL res=;bool neg=;char z;
        for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
        if(z==EOF)return ;
        if(z=='-'){neg=;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*+z-'',z=getchar());
        return (neg)?-res:res;
    }

const int maxN=;
const int maxM=*;

int N,K,M;
struct Troad
  {
      int x,y,id,p,cost;
        inline Troad(int _id=,int _p=,int _x=,int _y=,int _cost=){id=_id;p=_p;x=_x;y=_y;cost=_cost;}
  }road[maxM+];

inline bool cmpcost(Troad a,Troad b){return a.cost<b.cost;}

int pa[maxN+];
inline int findroot(int a){return pa[a]<?a:pa[a]=findroot(pa[a]);}
inline void uni(int a,int b)
  {
      int f1=findroot(a),f2=findroot(b);
      if(f1==f2)return;
      if(pa[f1]>pa[f2])swap(f1,f2);
      pa[f1]+=pa[f2];
      pa[f2]=f1;
  }

inline int check(int mid)
  {
      int i,cnt=;
      mmst(pa,-);
      re(i,,mid)
          if(road[i].p==)
            if(findroot(road[i].x)!=findroot(road[i].y))
              {
                  uni(road[i].x,road[i].y);
                  cnt++;
              }
        if(cnt<K)return ;
        re(i,,mid)uni(road[i].x,road[i].y);
        re(i,,N)if(findroot(i)!=findroot())return ;
        return ;
    }

struct Tout
  {
      int id,p;
      inline Tout(int _id=,int _p=){id=_id;p=_p;}
    }out[maxN+];
int ge;
inline bool cmpid(Tout a,Tout b){return a.id<b.id;}

int main()
  {
      freopen("road.in","r",stdin);
      freopen("road.out","w",stdout);
      int i;
      N=gint();K=gint();M=gint();
      re(i,,M)
        {
            int x=gint(),y=gint(),c1=gint(),c2=gint();
            road[*i-]=Troad(i,,x,y,c1);
            road[*i]=Troad(i,,x,y,c2);
        }
      M*=;
      sort(road+,road+M+,cmpcost);
      int l=,r=M,mid;
      while(l<=r)
          {
              mid=(l+r)>>;
                if(check(mid))r=mid-;else l=mid+;
            }
      cout<<road[l].cost<<endl;
      mmst(pa,-);
      re(i,,l)
          if(road[i].p==)
            if(findroot(road[i].x)!=findroot(road[i].y))
              {
                  uni(road[i].x,road[i].y);
                  out[++ge]=Tout(road[i].id,road[i].p);
              }
        re(i,,l)if(findroot(road[i].x)!=findroot(road[i].y))
          {
              uni(road[i].x,road[i].y);
                out[++ge]=Tout(road[i].id,road[i].p);
            }
        sort(out+,out+ge+,cmpid);
        re(i,,ge)cout<<out[i].id<<" "<<out[i].p<<endl;
        return ;
  }