UVa 1151 买还是建

1     int r = x;
2     while (r != p[r])  r = p[r];
3     int i = x, j;
4     while (p[i] != r) 
5     {
6         j = p[i];
7         p[i] = r;
8         i = j;
9     }

  1 #include<iostream> 
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<vector>
  5 using namespace std;
  6 
  7 const int maxn = 1000 + 5;
  8 
  9 int n, m, q, cnt;
 10 int p[maxn];
 11 vector<int> g[10];  //方案集合
 12 int c[10];          //方案价格
 13 
 14 //边
 15 struct node
 16 {
 17     int u;
 18     int v;
 19     int dist;
 20 }edge[maxn*maxn];
 21 
 22 //点
 23 struct node2
 24 {
 25     int x, y;
 26 }a[maxn];
 27 
 28 int find(int x)
 29 {
 30     //return p[x] == x ? x : find(p[x]);   用这个会超时
 31     //路径压缩
 32     int r = x;
 33     while (r != p[r])  r = p[r];
 34     int i = x, j;
 35     while (p[i] != r) 
 36     {
 37         j = p[i];
 38         p[i] = r;
 39         i = j;
 40     }
 41     return r;
 42 }
 43 
 44 bool cmp(node a, node b)
 45 {
 46     return a.dist < b.dist;
 47 }
 48 
 49 //计算距离平方和
 50 int cacl_dist(node2 a, node2 b)
 51 {
 52     return (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y);
 53 }
 54 
 55 void init()
 56 {
 57     for (int k = 1; k <= n; k++)   p[k] = k;
 58 }
 59 
 60 int Kruskal()
 61 {
 62     int num = 0;
 63     int ans = 0;
 64     for (int i = 0; i < cnt ; i++)
 65     {
 66         int x = find(edge[i].u);
 67         int y = find(edge[i].v);
 68         if (x != y)
 69         {
 70             p[x] = y;
 71             ans += edge[i].dist;
 72             num++;
 73         }
 74         if (num == n - 1)  return ans;
 75     }
 76     return ans;
 77 }
 78 
 79 
 80 void solve()
 81 {
 82     init();
 83     int ans = Kruskal();
 84     //二进制枚举方案
 85     for (int i = 0; i < (1 << q); i++)
 86     {
 87         init();
 88         int cost = 0;
 89         for (int j = 0; j < q; j++)
 90         {
 91             if (i & (1 << j))
 92             {
 93                 cost += c[j];
 94                 int x = find(g[j][0]);
 95                 for (int k = 1; k < g[j].size(); k++)
 96                 {    
 97                     int y = find(g[j][k]);
 98                     if (x != y)
 99                         p[y] = x;
100                 }
101             }
102         }
103         ans = min(cost + Kruskal(), ans);
104     }
105     printf("%d\n", ans);
106 }
107 
108 int main()
109 {
110     //freopen("D:\\txt.txt", "r", stdin);
111     int T, t, s, kase=0;
112     scanf("%d", &T);
113     while (T--)
114     {
115         if (++kase > 1)    printf("\n");
116         scanf("%d%d", &n, &q);
117 
118         //存储方案
119         for (int i = 0; i < q; i++)
120         {
121             g[i].clear();
122             scanf("%d", &t);
123             scanf("%d", &c[i]);
124             for (int j = 0; j < t; j++)
125             {
126                 scanf("%d", &s);
127                 g[i].push_back(s);    
128             }
129         }
130 
131         for (int i = 1; i <= n; i++)
132             scanf("%d%d", &a[i].x, &a[i].y);
133 
134         //存储边
135         cnt = 0;
136         for (int i = 1; i <= n;i++)
137         for (int j = i + 1; j <= n; j++)
138         {
139             edge[cnt].u = i;
140             edge[cnt].v = j;
141             edge[cnt].dist = cacl_dist(a[i], a[j]);
142             cnt++;
143         }
144         sort(edge, edge + cnt, cmp);
145         solve();
146     }
147     return 0;
148 }