hdu 1520 Anniversary party 基础树形DP 树的最大独立集

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7575    Accepted Submission(s): 3323

Problem Description

There is going to be a party to celebrate the 80-th Anniversary of the Ural State University. The University has a hierarchical structure of employees. It means that the supervisor relation forms a tree rooted at the rector V. E. Tretyakov. In order to make the party funny for every one, the rector does not want both an employee and his or her immediate supervisor to be present. The personnel office has evaluated conviviality of each employee, so everyone has some number (rating) attached to him or her. Your task is to make a list of guests with the maximal possible sum of guests' conviviality ratings.

 

Input

Employees are numbered from 1 to N. A first line of input contains a number N. 1 <= N <= 6 000. Each of the subsequent N lines contains the conviviality rating of the corresponding employee. Conviviality rating is an integer number in a range from -128 to 127. After that go T lines that describe a supervisor relation tree. Each line of the tree specification has the form: 
L K 
It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line 
0 0

 

Output

Output should contain the maximal sum of guests' ratings.

 

Sample Input

711111111 32 36 47 44 53 50 0

 

Sample Output

5

 

Source

Ural State University Internal Contest October'2000 Students Session

 

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题意：

给定一个树，选择若干点，使得选择的结点中任两点不相邻，求能选结点最大数量。同时判断方案数是否为一。

当初了两个while(~scanf())都都写掉了~，莫名其妙地发现TLE。

其实每组数据只会输入n个关系，其中最后一个关系都是0 0。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI (4.0*atan(1.0))#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   ind<<1,le,mid#define rson    ind<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     second#define ysk(x)  (1<<(x))using namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;const int maxn= 6000+20   ;//const int maxm=    ;int n,root;vector<int >G[maxn];int deg[maxn];int dp[maxn][2];int a[maxn];int dfs(int x,int se){    int &ans=dp[x][se];    if(ans!=-1)  return ans;    ans=0;    if(se)  ans+=a[x];    for(int i=0;i<G[x].size();i++)    {        int y=G[x][i];        if(se)  ans+=dfs(y,0);        else   ans+=max( dfs(y,0),dfs(y,1) );    }    return ans;}int main(){   while(~scanf("%d",&n))   {//       if(n==0)  {scanf("%d",&n);break;}       for(int i=1;i<=n;i++)       {           G[i].clear();           deg[i]=0;        scanf("%d",&a[i]);       }       int x,y;       while(~scanf("%d%d",&x,&y)&&(x||y))       {           G[y].push_back(x);           deg[x]++;       }       for(int i=1;i<=n;i++)       {           if(deg[i]==0)  root=i;       }       memset(dp,-1,(n+1) *sizeof dp[0]);       printf("%d\n",max( dfs(root,0),dfs(root,1)  )  );   }    return 0;}

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<climits>#include<queue>#include<vector>#include<map>#include<sstream>#include<set>#include<stack>#include<cctype>#include<utility>#pragma comment(linker, "/STACK:102400000,102400000")#define PI (4.0*atan(1.0))#define eps 1e-10#define sqr(x) ((x)*(x))#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)#define  lson   ind<<1,le,mid#define rson    ind<<1|1,mid+1,ri#define MID   int mid=(le+ri)>>1#define zero(x)((x>0? x:-x)<1e-15)#define mk    make_pair#define _f     first#define _s     second#define ysk(x)  (1<<(x))using namespace std;//const int INF=    ;typedef long long ll;//const ll inf =1000000000000000;//1e15;//ifstream fin("input.txt");//ofstream fout("output.txt");//fin.close();//fout.close();//freopen("a.in","r",stdin);//freopen("a.out","w",stdout);const int INF =0x3f3f3f3f;const int maxn= 6000+20   ;//const int maxm=    ;int n,root;vector<int >G[maxn];int deg[maxn];int dp[maxn][2];int a[maxn];int dfs(int x,int se){    int &ans=dp[x][se];    if(ans!=-1)  return ans;    ans=0;    if(se)  ans+=a[x];    for(int i=0;i<G[x].size();i++)    {        int y=G[x][i];        if(se)  ans+=dfs(y,0);        else   ans+=max( dfs(y,0),dfs(y,1) );    }    return ans;}int main(){   while(~scanf("%d",&n))   {//       if(n==0)  {scanf("%d",&n);break;}       for(int i=1;i<=n;i++)       {           G[i].clear();           deg[i]=0;        scanf("%d",&a[i]);       }       int x,y;       for(int i=1;i<=n;i++)       {           scanf("%d%d",&x,&y);           G[y].push_back(x);           deg[x]++;       }       for(int i=1;i<=n;i++)       {           if(deg[i]==0)  root=i;       }       memset(dp,-1,(n+1) *sizeof dp[0]);       printf("%d\n",max( dfs(root,0),dfs(root,1)  )  );   }    return 0;}

我想说的是，不用调用两次递归来求，一次递归即可。

在函数里面对dp[x][0]分别处理即可。