Description Let’s imagine how apple tree looks in binary computer
world. You’re right, it looks just like a binary tree, i.e. any
biparous branch splits up to exactly two new branches. We will
enumerate by integers the root of binary apple tree, points of
branching and the ends of twigs. This way we may distinguish different
branches by their ending points. We will assume that root of tree
always is numbered by 1 and all numbers used for enumerating are
numbered in range from 1 to N, where N is the total number of all
enumerated points. For instance in the picture below N is equal to 5.
Here is an example of an enumerated tree with four branches:2 5 \ / 3 4 \ /
1As you may know it’s not convenient to pick an apples from a tree when
there are too much of branches. That’s why some of them should be
removed from a tree. But you are interested in removing branches in
the way of minimal loss of apples. So your are given amounts of apples
on a branches and amount of branches that should be preserved. Your
task is to determine how many apples can remain on a tree after
removing of excessive branches.Input First line of input contains two numbers: N and Q ( 2 ≤ N ≤ 100;
1 ≤ Q ≤ N ? 1 ). N denotes the number of enumerated points in a tree.
Q denotes amount of branches that should be preserved. Next N ? 1
lines contains descriptions of branches. Each description consists of
a three integer numbers divided by spaces. The first two of them
define branch by it’s ending points. The third number defines the
number of apples on this branch. You may assume that no branch
contains more than 30000 apples.Output Output should contain the only number — amount of apples that
can be preserved. And don’t forget to preserve tree’s root ;-)
树形dp。dp[i][j]表示以i为根的子树中取j条边的最优解。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[110][110],fir[210],ne[210],to[210],len[210],n,m;
void add(int num,int f,int t,int l)
{ne[num]=fir[f];fir[f]=num;to[num]=t;len[num]=l;
}
void dfs(int u,int fa)
{int i,j,k,v,x,y,z;for (i=fir[u];i;i=ne[i])if (to[i]!=fa){v=to[i];dfs(v,u);for (j=m;j;j--)for (k=0;k<j;k++)dp[u][j]=max(dp[u][j],dp[u][j-k-1]+dp[v][k]+len[i]);}
}
int main()
{int i,j,k,p,q,x,y,z;scanf("%d%d",&n,&m);for (i=1;i<n;i++){scanf("%d%d%d",&x,&y,&z);add(i*2,x,y,z);add(i*2+1,y,x,z);}dfs(1,-1);printf("%d\n",dp[1][m]);
}