SPOJ COMPANYS Two Famous Companies 最小生成树,二分,思路 难度:2

#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=2e5+5;
typedef pair<int,int> P;
int first[maxn];
struct edge{
        int f,t,nxt,c;
        int tel;
}e[maxn],org[maxn];
int n,m,k,sum;
bool used[maxn];

void addedge(int f,int t,int c,int tel,int i){
        org[i].nxt=first[f];
        first[f]=i;
        org[i].f=f;
        org[i].t=t;
        org[i].c=c;
        org[i].tel=tel;
}
class Greater: public binary_function<P, P,bool>
{
public:
    bool operator()(const P& a, const P& b)
    {
        if(a.first!=b.first)return a.first>b.first;
        return e[a.second].tel<e[b.second].tel;
    }
};
priority_queue <P,vector<P>,Greater> pque;
int prim(){
    memset(used,false,sizeof(used));
    used[0]=true;
    int unum=1;
    for(int p=first[0];p!=-1;p=e[p].nxt){
        pque.push(P(e[p].c,p));
    }
    sum=0;
    int ans=0;
    while(unum<n&&!pque.empty()){
        int p=pque.top().second;
        int t=e[p].t;
        int td=pque.top().first;
        pque.pop();
        if(used[t])continue;
        if(e[p].tel)ans++;
        sum+=td;
        used[t]=true;
        unum++;

        for(int p=first[t];p!=-1;p=e[p].nxt){
                if(!used[e[p].t]){
                        pque.push(P(e[p].c,p));
                }
        }

    }
    while(!pque.empty())pque.pop();
    return ans;
}
int main(){
        int ki=0;
       while(scanf("%d%d%d",&n,&m,&k)==3&&++ki){
                memset(first,-1,sizeof(first));
                for(int i=0;i<m;i++){
                        int f,t,c,color;
                        scanf("%d%d%d%d",&f,&t,&c,&color);
                        addedge(f,t,c,1-color,2*i);
                        addedge(t,f,c,1-color,2*i+1);
                }
                int l=-110,r=110;
                int ans;
                while(l<=r){
                        int mid=(l+r)/2;
                        copy(org,org+2*m,e);
                        for(int i=0;i<2*m;i++){
                                if(e[i].tel){
                                        e[i].c+=mid;
                                }
                        }
                        int kk=prim();
                     //   if(kk==k){ans=sum-k*mid;break;}
                        if(kk<k)r=mid-1;
                        else {ans=sum-k*mid;l=mid+1;}
                }
                printf("Case %d: %d\n",ki,ans);
        }
        return 0;
}