Codeforces 815C. Karen and Supermarket 【树形DP】

C. Karen and Supermarket

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

On the way home, Karen decided to stop by the supermarket to buy some groceries.

She needs to buy a lot of goods, but since she is a student her budget is still quite limited. In fact, she can only spend up to b dollars.

The supermarket sells n goods. The i-th good can be bought for ci dollars. Of course, each good can only be bought once.

Lately, the supermarket has been trying to increase its business. Karen, being a loyal customer, was given n coupons. If Karen purchases the i-th good, she can use the i-th coupon to decrease its price by di. Of course, a coupon cannot be used without buying the corresponding good.

There is, however, a constraint with the coupons. For all i ≥ 2, in order to use the i-th coupon, Karen must also use the xi-th coupon (which may mean using even more coupons to satisfy the requirement for that coupon).

Karen wants to know the following. What is the maximum number of goods she can buy, without exceeding her budget b?

Input

The first line of input contains two integers n and b (1 ≤ n ≤ 5000, 1 ≤ b ≤ 109), the number of goods in the store and the amount of money Karen has, respectively.

The next n lines describe the items. Specifically:

• The i-th line among these starts with two integers, ci and di (1 ≤ di < ci ≤ 109), the price of the i-th good and the discount when using the coupon for the i-th good, respectively.
• If i ≥ 2, this is followed by another integer, xi (1 ≤ xi < i), denoting that the xi-th coupon must also be used before this coupon can be used.

Output

Output a single integer on a line by itself, the number of different goods Karen can buy, without exceeding her budget.

Examples

input

6 1610 910 5 112 2 120 18 310 2 32 1 5

output

4

input

5 103 13 1 13 1 23 1 33 1 4

output

5

Note

In the first test case, Karen can purchase the following 4 items:

• Use the first coupon to buy the first item for 10 - 9 = 1 dollar.
• Use the third coupon to buy the third item for 12 - 2 = 10 dollars.
• Use the fourth coupon to buy the fourth item for 20 - 18 = 2 dollars.
• Buy the sixth item for 2 dollars.

The total cost of these goods is 15, which falls within her budget. Note, for example, that she cannot use the coupon on the sixth item, because then she should have also used the fifth coupon to buy the fifth item, which she did not do here.

In the second test case, Karen has enough money to use all the coupons and purchase everything.

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显然,可以根据优惠券的约束关系构造成一颗树,考虑树形DP 

g[u][i]=u为根的子树,购买了i件物品,不能使用优惠卷的最小花销

f[u][i]=u为根的子树,买了i件物品,能使用优惠券,且必定用优惠价买了u物品的最小花销

f′[u][i]=f[u][i+1]−(c[u]−d[u])

---

g[u][i]=min{g[u][i],g[u][i−j]+g[v][j]} | (0<=j<=i,v为u的儿子节点)

f′[u][i]=min{f′[u][i],f′[u][i−j]+min(g[v][j],f[v][j])} | (0<=j<=i,v为u的儿子节点)

f[u][i]=f′[u][i−1]+(c[u]−d[u])

---

显然,直接DP需要O(n3) 
有个比较玄学的优化,维护size[u]=当前u为根的子树已经发现的节点个数 
首先对任意u,size[u]=1,求出儿子v后,背包问题搞一下,然后size[u]+=size[v],继续求解其他儿子... 
这样复杂度其实只有O(n2),因为子树u内任意2个点对,只同时求了一次 
注意如果size[u]+=size[v],再对u,v求背包,复杂度就变为O(n3)了,因为子树v也与自身求了一次背包

#include<stdio.h>#include<bits/stdc++.h>#define ll long long#define pii pair<int,int>#define MEM(a,x) memset(a,x,sizeof(a))#define lowbit(x) ((x)&-(x))using namespace std;const int inf=0x3f3f3f3f;const int N = 5e3+5;vector<int>G[N];int c[N],d[N],f[N][N],g[N][N],size[N];void init(){    MEM(f,0x3f);    MEM(g,0x3f);    MEM(size,0);    for(int i=0;i<N;++i){        G[i].clear();        f[i][0]=g[i][0]=0;    }}void dp(int u,int n){    g[u][1]=c[u];    size[u]=1;    for(auto v:G[u]){        dp(v,n);        //size[u]+=size[v]; 放这里O(n^3)        for(int i=size[u];i>=0;--i){            for(int j=size[v];j>=0;--j){                g[u][i+j]=min(g[u][i+j],g[u][i]+g[v][j]);                if(i!=0){                    f[u][i+j-1]=min(f[u][i-1+j],f[u][i-1]+min(g[v][j],f[v][j]));                }            }        }        size[u]+=size[v];//O(n^2)    }    for(int i=size[u];i>=1;--i){        f[u][i]=f[u][i-1]+c[u]-d[u];    }}int main(){    //freopen("/home/lu/code/r.txt","r",stdin);    int n,b;    while(~scanf("%d%d",&n,&b)){        init();        scanf("%d%d",&c[0],&d[0]);        for(int i=1;i<n;++i){            int x;            scanf("%d%d%d",&c[i],&d[i],&x);            G[x-1].push_back(i);        }        dp(0,n);        for(int i=n;i>=0;--i){            if(f[0][i]<=b||g[0][i]<=b){                printf("%d\n",i);                break;            }        }    }    return 0;}