Uva1151

来自：http://blog.youkuaiyun.com/a197p/article/details/48691245

Kurskal+二进制枚举子集的方法

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
struct point {
    int x,y;
}pp[maxn];
struct edge{
    int s,e,dist;
}l[maxn*maxn];
int q,n,m;
int p[maxn];
vector<int> g[10];
int c[10];
int distance_(point a,point b){
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int cmp(edge a,edge b){
    return a.dist < b.dist;
}
int find_(int x){
    return p[x]==x?x:p[x]=find_(p[x]);
}
bool merge_(int a,int b){
    int x=find_(a);
    int y=find_(b);
    if(x==y) return false;
    p[x]=y;
    return true;
}
int kruskal(){
    int ans=0;
    int num=0;
    for(int i=0;i<m&&num<n-1;i++){
        if(merge_(l[i].s,l[i].e)){
            num++;
            ans+=l[i].dist;
        }
    }
    return ans;
}
void solve(){
    for(int i=0;i<=n;i++) p[i]=i;
    int ans=kruskal();
    for(int s=1;s<(1<<q);s++){
        int cost=0;
        for(int tt=0;tt<=n;tt++) p[tt]=tt;
        for(int j=0;j<q;j++)
        {
            if(!((s>>j)&1)) continue;
            cost+=c[j];
            for(int k=0;k<g[j].size();k++)
            {
                merge_(g[j][k],g[j][0]);
            }
        }
        ans=min(ans,cost + kruskal());
    }
    printf("%d\n",ans);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&q);
        for(int i=0;i<10;i++) g[i].clear();
        for(int i=0;i<q;i++){
            int cnt=0;
            scanf("%d%d",&cnt,&c[i]);
            int a;
            for(int j=0;j<cnt;j++){
                scanf("%d",&a);
                g[i].push_back(a);
            }
        }
        for(int i=1;i<=n;i++){
            scanf("%d%d",&pp[i].x,&pp[i].y);
        }
        m=0;
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++)
            {
                l[m].s=i;
                l[m].e=j;
                l[m++].dist=distance_(pp[i],pp[j]);
            }
        }
        sort(l,l+m,cmp);
        solve();
        if(t) printf("\n");
    }
    return 0;
}