hdu1561--H - ACboy needs your help(树形dp)

H - ACboy needs your help

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Description

ACboy has N courses this term, and he plans to spend at most M days on study.Of course,the profit he will gain from different course depending on the days he spend on it.How to arrange the M days for the N courses to maximize the profit? 

 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers N and M, N is the number of courses, M is the days ACboy has. 
Next follow a matrix A[i][j], (1<=i<=N<=100,1<=j<=M<=100).A[i][j] indicates if ACboy spend j days on ith course he will get profit of value A[i][j]. 
N = 0 and M = 0 ends the input. 

 

Output

For each data set, your program should output a line which contains the number of the max profit ACboy will gain. 

 

Sample Input

 2 21 21 32 22 12 12 33 2 13 2 10 0 

 

Sample Output

 346 

 

树形dp第一题啊

题目给出的依赖关系，可以组成几个树，在虚拟一个root，节点0，这样把所有的点连接到一起

树形dp中，数组dp[i][j]表示，对于第i个节点来说，选取j个城市，可以达到的最大值，建树完成后，dfs，递归到叶子节点，dp[叶子][1] = 叶子的权值，其他的都是0，然后回溯上来，回到父节点，dp[i][j] = max( dp[i][j],dp[i][j-k]+dp[v][k] ) v表示孩子节点，k表示在整个以v为根节点的子树中选取k个城市，那么在以i为根节点的城市只能选取j-k个了，这样近似于将i的每一个子节点都看做分组背包中的一个组，进行分组背包的操作，来得到dp[i][j] （1<=j <= m）的最大的权值，然后逐层回溯。

注意1，增加的总的根root，所以要求的城市数m应该再加上1.

注意2，因为存在依赖关系，所以dp[i][1]一定是i节点的权值，之后的才是子节点可以改变的权值，所以dp[i][j],j应该大于等于2，而且k一定要小于j，留下的一个位置放第i个节点

 

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct node{    int v ;    int next ;} edge[1000000] ;int head[300] , p[300] , cnt ;int dp[300][300] ;int n , m ;void add(int u,int v){    edge[cnt].v = v ;    edge[cnt].next = head[u] ;    head[u] = cnt++ ;}void dfs(int u){    int i , v , j , k ;    dp[u][1] = p[u] ;    for(i = head[u] ; i != -1 ; i = edge[i].next)    {        v = edge[i].v ;        dfs(v);        for(j = m ; j >= 2 ; j--)        {            for(k = 0 ; k < j ; k++)                dp[u][j] = max( dp[u][j], dp[u][j-k]+dp[v][k] );        }    }}int main(){    int u , v , i ;    while( scanf("%d %d", &n, &m)!=EOF )    {        if(n == 0 && m == 0) break;        memset(head,-1,sizeof(head));        memset(dp,0,sizeof(dp));        p[0] = cnt = 0 ;        for(i = 1 ; i <= n ; i++)        {            scanf("%d %d", &u, &v);            p[i] = v ;            add(u,i);        }        m++ ;        dfs(0);        printf("%d\n", dp[0][m]);    }    return 0;}