BNUOJ-26482 Juice 树形DP

　　题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=26482

　　题意：给一颗树，根节点为送电站，可以无穷送电，其它节点为house，电量达到pi时可以点亮，边为电线，传输有容量上限，求最多点亮多少个house。。

　　简单树形DP，f[i][j]表示第 i 个节点电量为 j 时最多点亮的house个数。那么f[u][j]=Max{ f[u][j], f[u][j-k]+f[v][k] | v为u的儿子节点 }。。

 //STATUS:C++_AC_208MS_2032KB
 #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=1e9+,STA=;
 //const LL LNF=1LL<<60;
 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,w;
 }e[N];
 int first[N],next[N];
 int n,mt;
 int p[N],c[N],f[N][],vis[N];

 void adde(int a,int b,int c)
 {
     e[mt].u=a,e[mt].v=b;e[mt].w=c;
     next[mt]=first[a],first[a]=mt++;
 }

 void dfs(int u,int up)
 {
     int v,i,j,k;
     vis[u]=;
     f[u][p[u]]=;
     for(i=first[u];i!=-;i=next[i]){
         v=e[i].v;
         dfs(v,e[i].w);
         for(j=up;j>=;j--){
             for(k=;k<=j && k<=e[i].w;k++){
                 f[u][j]=Max(f[u][j],f[u][j-k]+f[v][k]);
             }
         }
     }
 }

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,k,a,b,c;
     while(~scanf("%d",&n))
     {
         mem(first,-);mt=;
         for(i=;i<=n;i++){
             scanf("%d%d%d",&a,&p[i],&c);
             adde(a,i,c);
         }

         mem(vis,);mem(f,);
         int ans=;
         for(i=first[];i!=-;i=next[i]){
             int v=e[i].v;
             dfs(v,e[i].w);
             int hig=;
             for(j=;j<=e[i].w;j++)hig=Max(hig,f[v][j]);
             ans+=hig;
         }
         for(i=;i<=n;i++){
             if(!vis[i] && !p[i])ans++;
         }

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