Winter Day10: dp + tree bipartite graph matching (Hungarian Algorithm) Exercise

Tree dp

dp[N][2]：

// dp for the current node is selected or not selected
// dp [i] [0] ; // not selected
// dp [i] [1] ; // selected from

State transition equation (the key):

for(int i=0;i<v[root].size();i++)
{
　　DP [the root] [ 0 ] + DP = [V [the root] [I]] [ . 1 ]; // if the current node is not selected, the sub-node must choose 
　　DP [the root] [ . 1 ] + = min (DP [V [the root] [I]] [ . 1 ], DP [V [the root] [I]] [ 0 ]); // current node is selected, a child node (selected / not selected) 
}

 

1  // given a node number n ~ n-1 0
 2  // each station a sentinel node, each sentry observed with their two adjacent sides
 3  // Q required minimum number of sentinel can be observed All edges 
. 4  
. 5 #include < String .h>
 . 6 #include <the iostream>
 . 7 #include <stdio.h>
 . 8 #include <algorithm>
 . 9 #include <Queue>
 10 #include <Vector>
 . 11 #include <Map>
 12 is #include <the cmath>
 13 is  the using  namespace STD;
 14  #define INF 0x3f3f3f3f
 15  #define inff 0x3f3f3f3f3f3f3f3f
 16  const int N = 1550 ;
 . 17  #define MOD 998 244 353
 18 is typedef Long  Long LL;
 . 19  
20 is  int n-;
 21 is Vector < int > V [N];
 22 is  int F [N], DP [N] [ 2 ];
 23 is  // DP for the current node is selected or not selected
 24  // DP [I] [0]; // no option
 25  // DP [I] [. 1]; // option 
26 is  void DFS ( int the root)
 27  {
 28      DP [the root] [ 0 ] =0 ; // this is not the root node is selected from 
29      DP [the root] [ . 1 ] = . 1 ; // currently selected root 
30      for ( int I = 0 ; I <V [the root] .size (); I ++ )
 31 is          DFS ( V [the root] [I]);
 32      for ( int I = 0 ; I <V [the root] .size (); I ++ )
 33 is      {
 34 is          DP [the root] [ 0 ] + = DP [V [the root] [I ]] [ 1 ]; // if the current node is not selected, the sub-node must be selected from the 
35          DP [the root] [ 1 ] + = min (DP [V [the root] [I]] [ 1], DP [V [the root] [I]] [ 0 ]); // current node is selected, a child node (selected / not selected) 
36      }
 37 [  }
 38 is  
39  int main ()
 40  {
 41 is      the while (~ Scanf ( " % D " , & n-))
 42 is      {
 43 is          Memset (F, - . 1 , the sizeof (F)); // since the node ID 0 ~ n--. 1 
44 is          for ( int I = . 1 ; I <= n-; I ++ )
 45          {
 46 is              int the root, Node, Son;
 47              Scanf (" % D: (% D) " , & the root, & Node);
 48              V [the root] .clear (); // remember Clear 
49              for ( int J = . 1 ; J <= Node; J ++ )
 50              {
 51 is                  Scanf ( " % D " , & Son);
 52 is                  V [the root] .push_back (Son); // unidirectional edge 
53 is                  F [Son] = the root;
 54 is              }
 55          }
 56 is          int X = 0 ;
 57 is          the while (F [X] ! = -. 1 )
 58              X = F [X]; // find the root of the whole tree 
59          DFS (X);
 60          int Minn = min (DP [X] [ 0 ], DP [X] [ . 1 ]);
 61 is          the printf ( " % D \ n- " , Minn);
 62 is      }
 63 is      return  0 ;
 64 }

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Bipartite graph matching (Hungarian Algorithm) Exercise