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=1010;
const int INF=0x3f3f3f3f;
const int MOD=1e9+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

struct Edge{
    int u,v,w;
}e[N];
int first[N],next[N];
int n,mt;
int p[N],c[N],f[N][110],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]=1;
    f[u][p[u]]=1;
    for(i=first[u];i!=-1;i=next[i]){
        v=e[i].v;
        dfs(v,e[i].w);
        for(j=up;j>=0;j--){
            for(k=0;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,-1);mt=0;
        for(i=1;i<=n;i++){
            scanf("%d%d%d",&a,&p[i],&c);
            adde(a,i,c);
        }

        mem(vis,0);mem(f,0);
        int ans=0;
        for(i=first[0];i!=-1;i=next[i]){
            int v=e[i].v;
            dfs(v,e[i].w);
            int hig=0;
            for(j=0;j<=e[i].w;j++)hig=Max(hig,f[v][j]);
            ans+=hig;
        }
        for(i=1;i<=n;i++){
            if(!vis[i] && !p[i])ans++;
        }

        printf("%d
",ans);
    }
    return 0;
}