NEFU 641 素数树(树形DP)

素数树

Time Limit 1000ms Memory Limit 65536K

description

Suppose there is a tree named A. All nodes of A have a weight v(0 < v < 4000000).Now, we will give the definition of "Prime Node".A node is a Prime Node if the following conditions are satisfied.The subtree of A whose root node is b will be marked as B. If all nodes in B have prime weights and b's weight is greater than other nodes', then b is a Prime Node.Now you are to calculate how many Prime Nodes are in A.

input

For each test case:The fist line will contains an integer n(1 <= n <= 10000) indicating the number of nodes in A,the root of A will always be node 1.The second line has n integers giving the weights of each node numbered from 1 to n.The following n-1 lines, each contains a pair of integers x and y indicating there is an edge between x and y, give the n-1 edges of A. Process to the end of file.

output

For each test case, print the number of Prime Nodes on a single line.Print a blank line after each test case, even after the last one.

sample_input

5 12 11 5 7 3 1 2 2 3 2 5 3 4

sample_output

3

hint

source

思路：记忆化搜索，一个状态中有四个元素，一个标记每个点是不是素数节点，一个标记该点有没被访问，一个点标明以该点为根的子树最大值，一个点标明点值。

就是用个CLASS就好了。

class node
{
  public:bool vis,pri;
  int no,ma;
}f[nn];

然后从，主根节点1开始记忆化搜索其连边，如果改点以访问就返回，不是素数那将根也改为非素数，因为以这棵树为子树的树都不合条件。

失误点：这题T了我好几次，忒蛋了，原因就是整素数的时候太马虎，把一个优化给落下了。要不比赛时就过了。

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<set>
#include<vector>
using namespace std;
const int mm=4e6+9;
const int nn=5e4+9;
class node
{
  public:bool vis,pri;
  int no,ma;
}f[nn];
int head[nn],ver[nn],next[nn],edge;
bool p[mm],vis[nn];
int n,ans,ma[nn];
void ppp()
{ memset(p,0,sizeof(p));
  p[0]=1;p[1]=1;
  for(int i=2;i*i<mm;++i)
  { if(!p[i])///就是这个if忘记加了，T了
    for(int j=i+i;j<mm;j+=i)
      p[j]=1;
  }
}
void data()
{ memset(head,-1,sizeof(head));
  memset(vis,0,sizeof(vis));
  edge=0;
}
void add(int u,int v)
{
  ver[edge]=v;next[edge]=head[u];head[u]=edge++;
  ver[edge]=u;next[edge]=head[v];head[v]=edge++;
}
inline void Max(int&x,const int y)
{
  if(x<y)x=y;
}
void dfs(int u)
{ int v;
  bool yes=f[u].pri;
  f[u].vis=1;
  for(int i=head[u];i>=0;i=next[i])
  {
    v=ver[i];
    if(!f[v].vis)///未访问
    {
      f[v].vis=1;
      dfs(v);///搜其子树
      if(!f[v].pri)///子树不是素数树，就更改根为非素数树
        {f[u].pri=0;yes=0;}
      else
      {if(f[v].ma>=f[u].no)yes=0;///是素数树，判断是否根值最大
       Max(f[u].ma,f[v].ma);///更新
      }
    }
  }
  if(yes)///是素数树
  { ++ans;
  }
}
int main()
{ ppp();
  int u,v;
  while(scanf("%d",&n)!=EOF)
  { data();
    for(int i=1;i<=n;++i)
      {scanf("%d",&f[i].no);
       if(!p[f[i].no])
       {
         f[i].pri=1;
       }
       else f[i].pri=0;
       f[i].vis=0;
       f[i].ma=f[i].no;
      }
     for(int i=1;i<n;++i)
     {
       scanf("%d%d",&u,&v);
       add(u,v);
     }
     ans=0;
     memset(vis,0,sizeof(vis));
     dfs(1);
     printf("%d\n\n",ans);
  }
}