ZOJ-3201 Tree of Tree 树形DP

　　题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3201

　　题意：给一颗树，每个节点有一个权值，求节点数为n的最大权子树。

　　任意选择一个节点为根，然后DP转移就可以了，类似于分组背包。。。

 //STATUS:C++_AC_0MS_324KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef long long LL;
 typedef unsigned long long ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=1e+,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End
 int first[N],next[N*],val[N];
 int f[N][N];
 int n,m,mt,ans;
 struct Edge{
     int u,v;
 }e[N*];
 void adde(int a,int b)
 {
     e[mt].u=a,e[mt].v=b;
     next[mt]=first[a],first[a]=mt++;
     e[mt].u=b,e[mt].v=a;
     next[mt]=first[b],first[b]=mt++;
 }
 void dfs(int u,int fa)
 {
     int i,j,k,v;
     f[u][]=val[u];
     for(i=first[u];i!=-;i=next[i]){
         v=e[i].v;
         if(v==fa)continue;
         dfs(v,u);
         for(j=m;j>;j--){
             for(k=;k<j;k++){
                 f[u][j]=Max(f[u][j],f[u][j-k]+f[v][k]);
             }
         }
     }
     ans=Max(ans,f[u][m]);
     return ;
 }
 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,a,b;
     while(~scanf("%d%d",&n,&m))
     {
         for(i=;i<n;i++){
             scanf("%d",&val[i]);
         }
         mem(first,-);mt=;
         for(i=;i<n;i++){
             scanf("%d%d",&a,&b);
             adde(a,b);
         }
         mem(f,);ans=;
         dfs(,-);
         printf("%d\n",ans);
     }
     return ;
 }