UVA 1151

链接：

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3592
题意&题解：

紫书P358
代码：

 31 int n, m;
 32 VI v[10];
 33 PII p[MAXN];
 34 struct Edge {
 35     int u, v, cost;
 36     bool operator<(const Edge& t) {
 37         return cost < t.cost;
 38     }
 39 };
 40 vector<Edge> E;
 41 vector<Edge> E1;
 42 int par[MAXN];
 43 
 44 int find(int x) {
 45     return par[x] = par[x] == x ? x : find(par[x]);
 46 }
 47 
 48 void unite(int x, int y) {
 49     int a = find(x), b = find(y);
 50     if (a != b) par[a] = b;
 51 }
 52 
 53 int kruskal() {
 54     sort(all(E));
 55     rep(i, 1, n + 1) par[i] = i;
 56     int res = 0;
 57     rep(i, 0, E.size()) {
 58         Edge e = E[i];
 59         if (find(e.u) != find(e.v)) {
 60             unite(e.u, e.v);
 61             res += e.cost;
 62             E1.pb(e);
 63         }
 64     }
 65     return res;
 66 }
 67 
 68 int  kruskal1() {
 69     int res = 0;
 70     rep(i, 0, E1.size()) {
 71         Edge e = E1[i];
 72         if (find(e.u) != find(e.v)) {
 73             unite(e.u, e.v);
 74             res += e.cost;
 75         }
 76     }
 77     return res;
 78 }
 79 
 80 void init() {
 81     rep(i, 0, 10) v[i].clear();
 82     E.clear();
 83     E1.clear();
 84 }
 85 
 86 int main() {
 87     ios::sync_with_stdio(false), cin.tie(0);
 88     int T;
 89     cin >> T;
 90     while (T--) {
 91         init();
 92         cin >> n >> m;
 93         rep(i, 0, m) {
 94             int k;
 95             cin >> k;
 96             int x;
 97             cin >> x;
 98             v[i].pb(x);
 99             while (k--) {
100                 cin >> x;
101                 v[i].pb(x);
102             }
103         }
104         rep(i, 1, n + 1) cin >> p[i].first >> p[i].second;
105         rep(i, 1, n + 1) rep(j, i + 1, n + 1)
106             E.pb(Edge{ i, j, (p[i].first - p[j].first)*(p[i].first - p[j].first)
107                 + (p[i].second - p[j].second)*(p[i].second - p[j].second) });
108         int ans = kruskal();
109         rep(i, 0, 1 << m) {
110             rep(j, 1, n + 1) par[j] = j;
111             int sum = 0;
112             rep(j, 0, m) if (i&(1 << j)) {
113                 sum += v[j][0];
114                 rep(k, 1, v[j].size()) unite(v[j][1], v[j][k]);
115             }
116             ans = min(ans, sum + kruskal1());
117         }
118         cout << ans << endl;
119         if (T) cout << endl;
120     }
121     return 0;
122 }