[codevs1418]愚蠢的矿工

题目←

之前听过思路，胡搞了一下……
左儿子右兄弟写起来简单些（枚举次数并不少）

转移：
dp[i][j]为到第i个点剩j人的最大可获价值
节点上留人则可由左儿子和右兄弟的和+自己转来；
节点上不留人只能由右兄弟转来

#include
#include
#include
#include
using namespace std;
const int MAXN = 200000 + 50;
int head[MAXN],next[MAXN << 1],tot;
struct edge
{
    int f,t;
}l[MAXN << 1];
int w[MAXN],a,b,L[MAXN],R[MAXN];
void init(int n)
{
    for(int i = 0;i <= n;i ++)
        head[i] = -1,R[i] = -1,L[i] = -1;
}
int n,m;
void build(int f,int t)
{
    l[++ tot] = (edge){f,t};
    next[tot] = head[f];
    head[f] = tot;
}

void solve(int u)
{
    if(head[u] == -1)return;
    int i = head[u];
    int p = l[i].t;
    L[u] = p;
    solve(p);
    for(i = next[i];i != -1;i = next[i])
    {
        R[p] = l[i].t;
        p = R[p];
        solve(p);
    }
}
int dp[1005][1005];
int dfs(int u,int sy)
{
    if(u == -1)return 0;
    if(!sy)return dp[u][sy] = 0;
    if(dp[u][sy])return dp[u][sy];
    int ans1 = 0,ans2 = 0;
    for(int i = 0;i <= sy - 1;i ++)
    {
        ans1 = max(ans1,dfs(L[u],i) + dfs(R[u],sy - 1 - i));
    }
    ans1 += w[u];
    for(int i = 0;i <= sy;i ++)
    {
        ans2 = max(ans2,dfs(R[u],sy - i));
    }
    return dp[u][sy] = max(ans1,ans2);
}
int main()
{
    scanf("%d%d",&n,&m);
    init(n);
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&w[i]);
    }
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d%d",&a,&b);
        build(a,b);
    }
    solve(0);
    printf("%d",dfs(L[0],m));
    return 0;
}