PKU 3164 Command Network 最小树形图

      题意：要求建立司令部到各个欧几里德平面上的节点，给定可建立的顶点对(u,v)  =  u 可建立单向信道至 v ，求司令部形成对所有节点的指挥需要的最小建设花费。
      算法：最小图形树，不解释~
 

 1 /*
 2 Problem: 3164        User: _mTy
 3 Memory: 872K        Time: 172MS
 4 Language: C++        Result: Accepted
 5
 6 Source Code
 7 */
 8 # include<cstdio>
 9 # include <cstring>
10 # include<cmath>
11 # define MAXN 120
12 # define inf 1000000000
13 typedef  double  elem_t;
14 elem_t edmonds(int n , elem_t mat[][MAXN * 2 ] , int *  pre);
15 int main(){
16     elem_t point[MAXN][ 2 ];
17     elem_t mat[MAXN * 2 ][MAXN * 2 ];
18     elem_t res , len;
19     int pre[MAXN];
20     int i , j , n , m , u , v;
21
22      while (scanf( " %d%d " ,& n ,& m) != EOF){
23          for (i = 0 ;i < n;i ++ )  for (j = 0 ;j < n;j ++ ) mat[i][j] = inf;
24          for (i = 0 ;i < n;i ++ ) scanf( " %lf%lf " ,& point[i][ 0 ] ,& point[i][ 1 ]);
25          for (i = 0 ;i < m;i ++ ){
26             scanf( " %d%d " ,& u ,& v);  -- u;  -- v;
27             len  =   pow (point[u][ 0 ] - point[v][ 0 ] , 2 ) + pow (point[u][ 1 ] - point[v][ 1 ] , 2 );
28
29             len  =   sqrt (len);
30             mat[u][v] = len;
31         }
32
33         memset(pre , 0 , sizeof (pre));
34         pre[ 0 ] =- 1 ;
35         res  =  edmonds(n , mat , pre);
36          if (res < 0 )  printf ( " poor snoopy\n " );
37          else   printf ( " %.2f\n " , res);
38     }
39      return   0 ;
40 }
41
42 // 多源最小树形图,edmonds算法,邻接阵形式,复杂度O(n^3)
43 //返回最小生成树的长度,构造失败返回负值
44 //传入图的大小n和邻接阵mat,不相邻点边权inf
45 //可更改边权的类型,pre[]返回树的构造,用父结点表示
46 //传入时pre[]数组清零,用-1标出源点
47
48 elem_t edmonds(int n , elem_t mat[][MAXN * 2 ] , int *  pre){
49     elem_t ret = 0 ;
50     int c[MAXN * 2 ][MAXN * 2 ] , l[MAXN * 2 ] , p[MAXN * 2 ] , m = n , t , i , j , k;
51      for  (i = 0 ;i < n;l[i] = i , i ++ );
52      do {
53         memset(c , 0 , sizeof (c)) , memset(p , 0xff , sizeof (p));
54          for  (t = m , i = 0 ;i < m;c[i][i] = 1 , i ++ );
55          for  (i = 0 ;i < t;i ++ )
56              if  (l[i] == i && pre[i] !=- 1 ){
57                  for  (j = 0 ;j < m;j ++ )
58                      if  (l[j] == j && i != j && mat[j][i] < inf && (p[i] ==- 1 || mat[j][i] < mat[p[i]][i]))
59                         p[i] = j;
60                  if  ((pre[i] = p[i]) ==- 1 )
61                      return   - 1 ;
62                  if  (c[i][p[i]]){
63                      for  (j = 0 ;j <= m;mat[j][m] = mat[m][j] = inf , j ++ );
64                      for  (k = i;l[k] != m;l[k] = m , k = p[k])
65                          for  (j = 0 ;j < m;j ++ )
66                              if  (l[j] == j){
67                                  if  (mat[j][k] - mat[p[k]][k] < mat[j][m])
68                                     mat[j][m] = mat[j][k] - mat[p[k]][k];
69                                  if  (mat[k][j] < mat[m][j])
70                                     mat[m][j] = mat[k][j];
71                             }
72                     c[m][m] = 1 , l[m] = m , m ++ ;
73                 }
74                  for  (j = 0 ;j < m;j ++ )
75                      if  (c[i][j])
76                          for  (k = p[i];k !=- 1 && l[k] == k;c[k][j] = 1 , k = p[k]);
77             }
78     }
79      while  (t < m);
80      for  (;m --> n;pre[k] = pre[m])
81          for  (i = 0 ;i < m;i ++ )
82              if  (l[i] == m){
83                  for  (j = 0 ;j < m;j ++ )
84                      if  (pre[j] == m && mat[i][j] == mat[m][j])
85                         pre[j] = i;
86                  if  (mat[pre[m]][m] == mat[pre[m]][i] - mat[pre[i]][i])
87                     k = i;
88             }
89      for  (i = 0 ;i < n;i ++ )
90          if  (pre[i] !=- 1 )
91             ret += mat[pre[i]][i];
92      return  ret;
93 }