HDU-4616 Game 树形DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4616

　　比较典型的树形DP题目，f[u][j][k]表示以点u为子树，经过 j 个陷阱的最大值，其中k=0表示从u点出发，k=1表示终点为点u。则转移方程为：f[u][j+is_rtap][k]=Max{ f[v][j][k] | v为u的儿子节点,0<=j<=m }。要分别对每颗子树求出最大值，枚举其中的两颗子树，一颗出，一颗进，更新最大值。要注意在更新k=0时的最大值是，j要从1开始遍历，因为如果当前u节点存在trap，而u中的其中一个子节点v没有trap，那么会使f[u][1][0]加上f[v][0][0]的值，而不满足题目中遇到满足的m后停止下来。

 //STATUS:C++_AC_93MS_5540KB
 #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>
 #pragma comment(linker,"/STACK:102400000,102400000")
 using namespace std;
 //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 __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=,M=;
 const int INF=0x3f3f3f3f;
 const int MOD=,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

 struct Edge{
     int u,v;
 }e[N<<];
 int first[N],next[N<<],tr[N];
 int T,n,m,mt;
 LL f[N][][],w[N],ans;

 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;
     for(i=first[u];~i;i=next[i]){
         v=e[i].v;
         if(v==fa)continue;
         dfs(v,u);
         for(j=;j<=m;j++){
             for(k=;k+j<=m;k++){
                 if(j<m)ans=Max(ans,f[u][j][]+f[v][k][]);
                 if(k<m)ans=Max(ans,f[u][j][]+f[v][k][]);
                 if(j+k<m)ans=Max(ans,f[u][j][]+f[v][k][]);
             }
         }
         for(j=;j+tr[u]<=m;j++){
             if(j && tr[u]<m)f[u][j+tr[u]][]=Max(f[u][j+tr[u]][],f[v][j][]+w[u]);
             if(j<m)f[u][j+tr[u]][]=Max(f[u][j+tr[u]][],f[v][j][]+w[u]);
         }
     }
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,a,b;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         mem(f,);
         for(i=;i<n;i++){
             scanf("%I64d%d",&w[i],&tr[i]);
             f[i][tr[i]][]=f[i][tr[i]][]=w[i];
         }
         mt=;mem(first,-);
         for(i=;i<n;i++){
             scanf("%d%d",&a,&b);
             adde(a,b);
         }

         ans=;
         dfs(,-);

         printf("%I64d\n",ans);
     }
     return ;
 }