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 }。。

  1 //STATUS:C++_AC_208MS_2032KB

  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=1010;

 37 const int INF=0x3f3f3f3f;

 38 const int MOD=1e9+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 

 58 struct Edge{

 59     int u,v,w;

 60 }e[N];

 61 int first[N],next[N];

 62 int n,mt;

 63 int p[N],c[N],f[N][110],vis[N];

 64 

 65 void adde(int a,int b,int c)

 66 {

 67     e[mt].u=a,e[mt].v=b;e[mt].w=c;

 68     next[mt]=first[a],first[a]=mt++;

 69 }

 70 

 71 void dfs(int u,int up)

 72 {

 73     int v,i,j,k;

 74     vis[u]=1;

 75     f[u][p[u]]=1;

 76     for(i=first[u];i!=-1;i=next[i]){

 77         v=e[i].v;

 78         dfs(v,e[i].w);

 79         for(j=up;j>=0;j--){

 80             for(k=0;k<=j && k<=e[i].w;k++){

 81                 f[u][j]=Max(f[u][j],f[u][j-k]+f[v][k]);

 82             }

 83         }

 84     }

 85 }

 86 

 87 int main(){

 88  //   freopen("in.txt","r",stdin);

 89     int i,j,k,a,b,c;

 90     while(~scanf("%d",&n))

 91     {

 92         mem(first,-1);mt=0;

 93         for(i=1;i<=n;i++){

 94             scanf("%d%d%d",&a,&p[i],&c);

 95             adde(a,i,c);

 96         }

 97 

 98         mem(vis,0);mem(f,0);

 99         int ans=0;

100         for(i=first[0];i!=-1;i=next[i]){

101             int v=e[i].v;

102             dfs(v,e[i].w);

103             int hig=0;

104             for(j=0;j<=e[i].w;j++)hig=Max(hig,f[v][j]);

105             ans+=hig;

106         }

107         for(i=1;i<=n;i++){

108             if(!vis[i] && !p[i])ans++;

109         }

110 

111         printf("%d\n",ans);

112     }

113     return 0;

114 }