DP：树DP

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非常喜欢玩一种战略游戏，在一个地图上，有N座城堡。每座城堡都有一定的宝物，在每次游戏中ACboy同意攻克M个城堡并获得里面的宝物。但因为地理位置原因。有些城堡不能直接攻克，要攻克这些城堡必须先攻克其它某一个特定的城堡。你能帮ACboy算出要获得尽量多的宝物应该攻克哪M个城堡吗？

 

Input

每一个測试实例首先包含2个整数，N,M.(1 <= M <= N <= 200);在接下来的N行里。每行包含2个整数。a,b. 在第 i 行，a 代表要攻克第 i 个城堡必须先攻克第 a 个城堡，假设 a = 0 则代表能够直接攻克第 i 个城堡。

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

 

Output

对于每一个測试实例。输出一个整数。代表ACboy攻克M个城堡所获得的最多宝物的数量。

 

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

 

Sample Output

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
",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
",max(dp[root][0],dp[root][1]));
    }
    return 0;
}