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\n",ans); 104 } 105 return 0; 106 }
