DP: tree DP

Hello everyone , I meet you again , I'm the king of the whole stack , I've prepared for you today Idea Registration code .

The more, The Better

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5414 Accepted Submission(s): 3217

Problem Description

ACboy I like playing a strategy game very much , On a map , Yes N A castle . Every castle has a certain treasure , In every game ACboy Agree to conquer M Build a castle and get the treasures inside . But because of the geographical location . Some castles cannot be conquered directly , To conquer these castles, you must first conquer a specific Castle . You can help ACboy Figure out which treasure to conquer if you want to get as many as possible M A castle ？

Input

Each test case first contains 2 It's an integer ,N,M.(1 <= M <= N <= 200); In the following N line . Each row contains 2 It's an integer .a,b. In the i That's ok ,a It means to conquer the i A castle must be conquered first a A castle , hypothesis a = 0 Then it means that you can directly overcome the i A castle .

b On behalf of the i The number of treasures in a castle , b >= 0. When N = 0, M = 0 End of input .

Output

For each test case . Output an integer . representative ACboy Conquered M The number of treasures obtained by the castle .

Sample Input

    3 2
0 1
0 2
0 3
7 4
2 2
0 1
0 4
2 1
7 1
7 6
2 2
0 0

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Sample Output

    5
13
   
   dp[i][j] Said to i Root node j Maximum value of child nodes .#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<vector>
typedef long long LL;
using namespace std;
const int maxn=220;
int v[maxn];
int n,m;
int dp[maxn][maxn];
vector<int>s[maxn];
void tree_dp(int n,int f)
{
    int len=s[n].size();
    dp[n][1]=v[n];
    for(int i=0;i<len;i++)
    {
        if(f>1)  tree_dp(s[n][i],f-1);
        for(int j=f;j>=1;j--)
        {
            for(int k=1;k<=j;k++)
                dp[n][j+1]=max(dp[n][j+1],dp[n][j+1-k]+dp[s[n][i]][k]);
        }
    }
}
int main()
{
    int f;
    while(~scanf("%d%d",&n,&m)&&(n+m))
    {
        v[0]=0;
        memset(dp,0,sizeof(dp));
        for(int i=0;i<=n;i++)
            s[i].clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&f,&v[i]);
            s[f].push_back(i);
        }
        tree_dp(0,m+1);
        printf("%d\n",dp[0][m+1]);
    }
    return 0;
}





    Anniversary party




Time Limit: 1000MS
 
Memory Limit: 65536K


Total Submissions: 4329
 
Accepted: 2463




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 N – 1 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
7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0

Sample Output
5

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
const int maxn=6005;
int dp[maxn][2],pre[maxn];
int visit[maxn],n;
void tree_dp(int x)
{
    visit[x]=1;
    for(int i=1;i<=n;i++)
    {
//        cout<<"111  "<<i<<endl;
        if(!visit[i]&&pre[i]==x)
        {
            tree_dp(i);
            dp[x][1]+=dp[i][0];
            dp[x][0]+=max(dp[i][1],dp[i][0]);
        }
    }
}

int main()
{
    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof(dp));
        memset(visit,0,sizeof(visit));
        memset(pre,0,sizeof(pre));
        for(int i=1;i<=n;i++)
           scanf("%d",&dp[i][1]);
        int x,y,root;
        while(~scanf("%d%d",&x,&y)&&(x+y))
        {
            pre[x]=y;
            root=y;
        }
        while(pre[root])
            root=pre[root];
 //       cout<<"fuck   "<<root<<endl;
        tree_dp(root);
        printf("%d\n",max(dp[root][0],dp[root][1]));
    }
    return 0;
}

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