b 代表第 i 个城堡的宝物数量, b >= 0。当N = 0, M = 0输入结束。

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

5

13

dp[i][j]表示以i为根节点j个子节点的最大值。
#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;

}