HDU -- 1520 -- tree dp <deep search is written>

I certainly haven't understood the tree-like DP yet... why does it feel like there is an extra deep search for the state transition equation? Or is this because tree-like DP relies on the tree-like data structure?

This topic is regarded by many people as a tree-like DP entry question indeed ....

If u is the precursor of V, that is, the parent

DP [u] [0] + = max (DP [v] [1], DP [v] [0]) 0 indicates that u is not included in 1

DP [U [[1] + = DP [v] [0]

Not hard to write.

The root node must be located for each tree-like DP.

Because the data is very loose, I directly use vector to write.

 1 #include <iostream> 2 #include <cstring> 3 #include <vector> 4 #include <algorithm> 5 using namespace std; 6  7 const int size = 6010; 8 vector<int>ve[size]; 9 int val[size];10 int dp[size][2];11 int in[size];12 13 void dfs( int u )14 {15     int Size , v;16     Size = ve[u].size();17     dp[u][1] = val[u];18     for( int i = 0 ; i<Size ; i++ )19     {20         v = ve[u][i];21         dfs(v);22         dp[u][0] += max( dp[v][0] , dp[v][1] );23         dp[u][1] += dp[v][0];24     }25 }26 27 int main()28 {29     cin.sync_with_stdio(false);30     int n , x , y , z;31     while( cin >> n )32     {33         memset( dp , 0 , sizeof(dp) );34         memset( in , 0 , sizeof(in) );35         for( int i = 1 ; i<=n ; i++ )36         {37             cin >> val[i];38             ve[i].clear();39         }40         while(1)41         {42             cin >> x >> y;43             ve[y].push_back(x);44             in[x] ++;45             if( !x && !y )46                 break;47         }48         for( int i =1 ; i<=n ; i++ )49         {50             if( !in[i] )51             {52                 z = i;53                 break;54             }55         }    56         dfs(z);57         cout << max( dp[z][0] , dp[z][1] ) << endl;58     }59     return 0;60 }

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HDU -- 1520 -- tree dp <deep search is written>