Red/Blue Spanning Tree

3 3 2
B 1 2
B 2 3
R 3 1
2 1 1
R 1 2
0 0 0

1
0

设红边为2，蓝边为1，求MST得最多蓝边数

同理得最少蓝边数，观察k是否在区间中

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
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 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 Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (3*2000+10)
#define MAXM (10000000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
class bingchaji
{
public:
    int father[MAXN],n;
    void mem(int _n)
    {
        n=_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)
    {
        father[x]=getfather(father[x]);
        father[y]=getfather(father[y]);
        father[father[x]]=father[father[y]];
    }
    bool same(int x,int y)
    {
        return getfather(x)==getfather(y);
    }
}S;

int n,m,k;
struct edge
{
    int a,b,c;
    edge(){}
    edge(int _x,int _y,int _c):a(_x),b(_y),c(_c){}
    friend bool operator<(edge a,edge b){return a.c<b.c;}
}e[MAXM];

int kr()
{
    S.mem(n);
    sort(e+1,e+1+m);
    int ans1=0,t=0;
    For(i,m) if (!S.same(e[i].a,e[i].b))
    {
        S.unite(e[i].a,e[i].b),ans1+=e[i].c,++t;
    //    cout<<e[i].a<<' '<<e[i].b<<' '<<e[i].c<<endl;
    }
    if (t<n-1) return -1;

    return ans1;
}
int main()
{
    //    freopen("g.in","r",stdin);
    //    freopen(".out","w",stdout);

    while(cin>>n>>m>>k)
    {
        if (n+m+k==0) return 0;
        S.mem(n);

        For(i,m)
        {
            char c;int a,b,t;
            scanf("\n%c %d %d",&c,&a,&b);
            if (c=='B') t=1;else t=2;
            e[i]=edge(a,b,t);
        }


        int t1=kr();

        For(i,m) e[i].c=3-e[i].c;

        int t2=kr();



//        cout<<t1<<' '<<t2<<endl;

        if (t1!=-1)
        {

            t1=t1-(n-1);
            t1=n-1-t1;
            t2=t2-(n-1);
  //          cout<<t1<<' '<<t2<<endl;

            if (t1>=k&&t2<=k)
                {
                    cout<<"1\n";
                    continue;
                }
        }


        cout<<"0\n";






    }

    return 0;
}

HDU 4263(Red/Blue Spanning Tree-取边贪心)

原文：http://blog.csdn.net/nike0good/article/details/47448389