• ZOJ-3201 Tree of Tree 树形DP


      题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3201

      题意:给一颗树,每个节点有一个权值,求节点数为n的最大权子树。

      任意选择一个节点为根,然后DP转移就可以了,类似于分组背包。。。

      1 //STATUS:C++_AC_0MS_324KB
      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=110;
     37 const int INF=0x3f3f3f3f;
     38 const int MOD=1e+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 int first[N],next[N*2],val[N];
     58 int f[N][N];
     59 int n,m,mt,ans;
     60 struct Edge{
     61     int u,v;
     62 }e[N*2];
     63 void adde(int a,int b)
     64 {
     65     e[mt].u=a,e[mt].v=b;
     66     next[mt]=first[a],first[a]=mt++;
     67     e[mt].u=b,e[mt].v=a;
     68     next[mt]=first[b],first[b]=mt++;
     69 }
     70 void dfs(int u,int fa)
     71 {
     72     int i,j,k,v;
     73     f[u][1]=val[u];
     74     for(i=first[u];i!=-1;i=next[i]){
     75         v=e[i].v;
     76         if(v==fa)continue;
     77         dfs(v,u);
     78         for(j=m;j>0;j--){
     79             for(k=0;k<j;k++){
     80                 f[u][j]=Max(f[u][j],f[u][j-k]+f[v][k]);
     81             }
     82         }
     83     }
     84     ans=Max(ans,f[u][m]);
     85     return ;
     86 }
     87 int main()
     88 {
     89  //   freopen("in.txt","r",stdin);
     90     int i,j,a,b;
     91     while(~scanf("%d%d",&n,&m))
     92     {
     93         for(i=0;i<n;i++){
     94             scanf("%d",&val[i]);
     95         }
     96         mem(first,-1);mt=0;
     97         for(i=1;i<n;i++){
     98             scanf("%d%d",&a,&b);
     99             adde(a,b);
    100         }
    101         mem(f,0);ans=0;
    102         dfs(0,-1);
    103         printf("%d
    ",ans);
    104     }
    105     return 0;
    106 }
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  • 原文地址:https://www.cnblogs.com/zhsl/p/3578769.html
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