H - ACboy needs your help
Time Limit:1000MS
Memory Limit:32768KB
64bit IO Format:%I64d & %I64u Submit Status
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 2
1 2
1 3
2 2
2 1
2 1
2 3
3 2 1
3 2 1
0 0
Sample Output
3
4
6
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
#include
#include
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; }