兩道相似的樹形dp+有依賴的的揹包,有依賴的揹包的多層依賴關係可以用樹表示出來,所以以後遇到多層依賴關係的揹包可以考慮將依賴關係表示成一棵樹,然後樹形DP求解,有依賴的揹包的求解使用到分組揹包思想,詳見揹包九講相關專題#include<iostream>
#include<cstdio>
#include<cstring>
#define M 210
using namespace std;
int n,m;
int head[M];
struct node{
int to,next;
}edge[M*2];
int dp[M][M];
int tot;
void insert(int u,int v){
edge[tot].to=v;
edge[tot].next=head[u];
head[u]=tot++;
}
void dfs(int u,int pre){
int i,j,k,v;
for(i=head[u];i!=-1;i=edge[i].next){
v=edge[i].to;
if(v==pre) continue;
dfs(v,u);
for(j=m;j>=1;j--)
for(k=1;k<=j-1;k++)
dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[v][k]);
}
}
int main(){
int i,j;
int u,val;
while(scanf("%d%d",&n,&m)&&n!=0&&m!=0){
tot=0;
m++;
memset(head,-1,sizeof(head));
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++){
scanf("%d%d",&u,&val);
insert(u,i);
insert(i,u);
for(j=1;j<=m;j++)
dp[i][j]=val;
}
dfs(0,-1);
printf("%d\n",dp[0][m]);
}
return 0;
}
#include<iostream>
#include<cstdio>
#define M 110
using namespace std;
int dp[M][M],num[M],head[M];
struct node{
int to,next;
}edge[M*2];
int n,m,tot;
void insert(int u,int v){
edge[tot].to=v;
edge[tot].next=head[u];
head[u]=tot++;
}
void dfs(int u,int pre){
int i,j,k,v;
for(i=head[u];i!=-1;i=edge[i].next){
v=edge[i].to;
if(v==pre) continue;
dfs(v,u);
for(j=m;j>=num[u];j--)
for(k=1;k<=j-num[u];k++)
dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[v][k]);
}
}
int main(){
int i,j;
int waste,value;
int from,to;
while(scanf("%d%d",&n,&m)&&n!=-1&&m!=-1){
tot=0;
memset(dp,0,sizeof(dp));
memset(head,-1,sizeof(head));
for(i=1;i<=n;i++){
scanf("%d%d",&waste,&value);
num[i]=(waste+19)/20;
for(j=num[i];j<=m;j++)
dp[i][j]=value;
}
for(i=1;i<n;i++){
scanf("%d%d",&from,&to);
insert(from,to);
insert(to,from);
}
if(m==0){
printf("0\n");
continue;
}
dfs(1,-1);
printf("%d\n",dp[1][m]);
}
return 0;
}