链接：

UVa http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3592

POJ http://poj.org/problem?id=2784

题意：告诉你n个点的坐标，建立一颗最小生成树，不过有q个套餐，套餐是连通某些点，并有一定花费，求最小生成树。

思路：n个点，n最大为1000，则最多有1000*999/2条边，先不使用套餐求一遍最小生成树，保留下最小生成树的边，此时最多有999条边，然后枚举每种套餐选或者不选，因为只有两种情况，用二进制枚举就行了，然后从筛选出的最多999条边里选边构造最小生成树。

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 5000100
#define eps 1e-7
#define INF 0x7FFFFFFF
#define LLINF 0x7FFFFFFFFFFFFFFF
#define seed 131
#define MOD 1000000007
#define ll long long
#define ull unsigned ll
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

struct node{
    int u,v;
    int w;
}edge[MAXN],edge2[1010];
int n,m,q,ans;
int qc[10][1010],qn[10],qm[10];
int x[1010],y[1010],father[1010];
bool cmp(node x,node y){
    return x.w<y.w;
}
int Find(int x){
    int t = x;
    while(t!=father[t])
        t = father[t];
    int k = x;
    while(k!=t){
        int temp = father[k];
        father[k] = t;
        k = temp;
    }
    return t;
}
void init(){
    for(int i=0;i<=n;i++)   father[i] = i;
}
int kruskal(){
    int i,j;
    int sum = 0;
    int tot = 0;
    for(i=0;i<m;i++){
        int a = Find(edge[i].u);
        int b = Find(edge[i].v);
        if(a!=b){
            father[a] = b;
            edge2[sum] = edge[i];
            tot += edge[i].w;
            sum++;
            if(sum>=n-1)    break;
        }
    }
    return tot;
}
int gao(int x){
    int i,j;
    int sum = 0;
    int tot = 0;
    for(i=0;i<q;i++){
        if(x&(1<<i)){
            tot += qm[i];
            for(j=0;j<qn[i]-1;j++){
                int aa = Find(qc[i][j]);
                int bb = Find(qc[i][j+1]);
                if(aa!=bb){
                    father[aa] = bb;
                    sum++;
                }
            }
        }
    }
    for(i=0;i<n-1;i++){
        int aa = Find(edge2[i].u);
        int bb = Find(edge2[i].v);
        if(aa!=bb){
            father[aa] = bb;
            sum++;
            tot += edge2[i].w;
            if(sum>=n-1)    break;
        }
    }
    //cout<<tot<<endl;
    return tot;
}
int main(){
    int t,i,j;
    scanf("%d",&t);
    for(int iii=0;iii<t;iii++){
        if(iii) puts("");
        m = 0;
        scanf("%d%d",&n,&q);
        for(i=0;i<q;i++){
            scanf("%d%d",&qn[i],&qm[i]);
            for(j=0;j<qn[i];j++){
                scanf("%d",&qc[i][j]);
            }
        }
        for(i=1;i<=n;i++){
            scanf("%d%d",&x[i],&y[i]);
            for(j=1;j<i;j++){
                int temp = (x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
                edge[m].u = i;
                edge[m].v = j;
                edge[m].w = temp;
                m++;
            }
        }
        sort(edge,edge+m,cmp);
        init();
        ans = kruskal();
        for(i=0;i<(1<<q);i++){
            init();
            int temp = gao(i);
            ans = min(ans,temp);
        }
        printf("%d\n",ans);
    }
    return 0;
}

UVa1151&POJ2784--Buy or Build【kruskal+二进制枚举】