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。。。

  1 //STATUS:C++_AC_0MS_324KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef long long LL;
 34 typedef unsigned long long ULL;
 35 //const
 36 const int N=110;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=1e+7,STA=8000010;
 39 const LL LNF=1LL<<60;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 int first[N],next[N*2],val[N];
 58 int f[N][N];
 59 int n,m,mt,ans;
 60 struct Edge{
 61     int u,v;
 62 }e[N*2];
 63 void adde(int a,int b)
 64 {
 65     e[mt].u=a,e[mt].v=b;
 66     next[mt]=first[a],first[a]=mt++;
 67     e[mt].u=b,e[mt].v=a;
 68     next[mt]=first[b],first[b]=mt++;
 69 }
 70 void dfs(int u,int fa)
 71 {
 72     int i,j,k,v;
 73     f[u][1]=val[u];
 74     for(i=first[u];i!=-1;i=next[i]){
 75         v=e[i].v;
 76         if(v==fa)continue;
 77         dfs(v,u);
 78         for(j=m;j>0;j--){
 79             for(k=0;k<j;k++){
 80                 f[u][j]=Max(f[u][j],f[u][j-k]+f[v][k]);
 81             }
 82         }
 83     }
 84     ans=Max(ans,f[u][m]);
 85     return ;
 86 }
 87 int main()
 88 {
 89  //   freopen("in.txt","r",stdin);
 90     int i,j,a,b;
 91     while(~scanf("%d%d",&n,&m))
 92     {
 93         for(i=0;i<n;i++){
 94             scanf("%d",&val[i]);
 95         }
 96         mem(first,-1);mt=0;
 97         for(i=1;i<n;i++){
 98             scanf("%d%d",&a,&b);
 99             adde(a,b);
100         }
101         mem(f,0);ans=0;
102         dfs(0,-1);
103         printf("%d\n",ans);
104     }
105     return 0;
106 }