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 ;

 }