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=110;
const int INF=0x3f3f3f3f;
const int MOD=1e+7,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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*2],val[N];
int f[N][N];
int n,m,mt,ans;
struct Edge{
    int u,v;
}e[N*2];
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][1]=val[u];
    for(i=first[u];i!=-1;i=next[i]){
        v=e[i].v;
        if(v==fa)continue;
        dfs(v,u);
        for(j=m;j>0;j--){
            for(k=0;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=0;i<n;i++){
            scanf("%d",&val[i]);
        }
        mem(first,-1);mt=0;
        for(i=1;i<n;i++){
            scanf("%d%d",&a,&b);
            adde(a,b);
        }
        mem(f,0);ans=0;
        dfs(0,-1);
        printf("%d
",ans);
    }
    return 0;
}