Kruskal 2015百度之星初赛2 HDOJ 5253 连接的管道

 1 /*
 2     最小生成树(Kruskal)：以权值为头，带入两个端点，自然的排序;感觉结构体的并查集很好看
 3     注意：题目老头要的是两个农田的高度差，中文水平不好，题意理解成和平均值的高度差！
 4 */
 5 #include <cstdio>
 6 #include <algorithm>
 7 #include <cstring>
 8 #include <cmath>
 9 #include <vector>
10 #include <queue>
11 #include <set>
12 using namespace std;
13 
14 const int MAXN = 1e3 + 10;
15 const int MAXM = 1e2 + 10;
16 const int INF = 0x3f3f3f3f;
17 int n, m;
18 int a[MAXN][MAXN];
19 vector<pair<int, int> > G[MAXM];
20 struct UF
21 {
22     int rt[MAXN*MAXN];
23     void init(void)    {memset (rt, -1, sizeof (rt));}
24 
25     int Find(int x)    {return (rt[x] == -1) ? x : rt[x] = Find (rt[x]);}
26 
27     void Union(int x, int y)
28     {
29         x = Find (x);    y = Find (y);
30         if (x > y)    rt[y] = x;
31         else if (x < y)    rt[x] = y;
32     }
33 
34     bool same(int x, int y)    {return Find (x) == Find (y);}
35 }uf;
36 
37 int Kruskal(void)
38 {
39     uf.init ();
40     int ans = 0;
41     for    (int i=0; i<=100; ++i)
42     {
43         for (int j=0; j<G[i].size (); ++j)
44         {
45             int u = G[i][j].first;    int v = G[i][j].second;
46             if (!uf.same (u, v))    {uf.Union (u, v);    ans += i;}
47         }
48     }
49 
50     return ans;
51 }
52 
53 int get_id(int i, int j)    {return (i - 1) * m + j;}
54 
55 int main(void)        //2015百度之星初赛2 HDOJ 5253 连接的管道
56 {
57     int t, cas = 0;    scanf ("%d", &t);
58     while (t--)
59     {
60         scanf ("%d%d", &n, &m);
61         for (int i=0; i<=100; ++i)    G[i].clear ();
62 
63         for (int i=1; i<=n; ++i)
64         {
65             for (int j=1; j<=m; ++j)
66             {
67                 scanf ("%d", &a[i][j]);
68                 if (i > 1)    {G[abs (a[i][j] - a[i-1][j])].push_back (make_pair (get_id (i, j), get_id (i-1, j)));}
69                 if (j > 1)    {G[abs (a[i][j] - a[i][j-1])].push_back (make_pair (get_id (i, j), get_id (i, j-1)));}
70             }
71         }
72 
73         printf ("Case #%d:\n", ++cas);
74         printf ("%d\n", Kruskal ());
75     }
76 
77     return 0;
78 }
79 
80 
81 /*
82 2
83 4  3
84 9 12 4
85 7 8 56
86 32 32 43
87 21 12 12
88 2  3
89 34 56 56
90 12 23 4
91 */