HDOJ 1561-tree DP, generalized backpack

Just read the question... I think this is not a tree... there may be loops... think twice .. this is really a tree (forest )... this is because each castle needs to break down at most one castle in advance .. for a castle. the castle that needs to be broken in advance as its father's diagram ....

Dp [k] [I] represents the maximum benefit that a k-followed subtree can achieve when it breaks through the I castle... generalized knapsack problem...

 

Program:

 

#include<iostream>#include<stack>#include<queue>#include<stdio.h>#include<algorithm>#include<string.h>#include<cmath>#define ll long long#define oo 1000000007#define MAXN 205using namespace std;vector<int> Tree[MAXN];int n,m,v[MAXN],ans[MAXN],dp[MAXN][MAXN];bool root[MAXN];void dfs(int x,int t){      int i,j,k,num;      if (t>m) return;      dp[x][1]=v[x];       num=Tree[x].size();      for (i=0;i<num;i++)      {            dfs(Tree[x][i],t+1);             for (j=m;j>=1;j--)               for (k=1;k<=m-j;k++)                  dp[x][j+k]=max(dp[x][j+k],dp[x][j]+dp[Tree[x][i]][k]);      }      return;}int main(){      int i,j,k;       while (~scanf("%d%d",&n,&m) && n && m)      {             for (i=1;i<=n;i++) Tree[i].clear();             memset(root,true,sizeof(root));             for (i=1;i<=n;i++)             {                   int x,c;                   scanf("%d%d",&x,&c);                   Tree[x].push_back(i);                   v[i]=c;                   if (x) root[i]=false;             }             memset(dp,0,sizeof(dp));             memset(ans,0,sizeof(ans));             for (i=1;i<=n;i++)               if (root[i])               {                      dfs(i,1);                      for (j=m;j>=0;j--)                         for (k=0;k<=m-j;k++)                             ans[j+k]=max(ans[j+k],ans[j]+dp[i][k]);               }            for (i=m;i>=0;i--)              if (ans[i]) break;            printf("%d\n",ans[i]);      }      return 0;}