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 ;
}