传送门 - > (CF816E) Karen and Supermarket
题意翻译
在回家的路上，凯伦决定到超市停下来买一些杂货。 她需要买很多东西，但因为她是学生，所以她的预算仍然很有限。
事实上，她只花了b美元。
超市出售N种商品。第i件商品可以以ci美元的价格购买。当然，每件商品只能买一次。
最近，超市一直在努力促销。凯伦作为一个忠实的客户，收到了n张优惠券。
如果凯伦购买i次商品，她可以用i次优惠券降低di美元。 当然，不买对应的商品，优惠券不能使用。
然而，对于优惠券有一个规则。对于所有i>=2，为了使用i张优惠券，凯伦必须也使用第xi张优惠券 （这可能意味着使用更多优惠券来满足需求。）
凯伦想知道。她能在不超过预算B的情况下购买的最大商品数量是多少？
感谢@luv_letters 提供的翻译
题目描述
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 bb dollars.
The supermarket sells nn goods. The ii -th good can be bought for c_{i}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 nncoupons. If Karen purchases the ii -th good, she can use the ii -th coupon to decrease its price by d_{i}di . Of course, a coupon cannot be used without buying the corresponding good.
There is, however, a constraint with the coupons. For all i>=2i>=2 , in order to use the ii -th coupon, Karen must also use the x_{i}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 bb ?
输入输出格式
输入格式：
The first line of input contains two integers nn and bb ( 1<=n<=50001<=n<=5000 , 1<=b<=10^{9}1<=b<=109 ), the number of goods in the store and the amount of money Karen has, respectively.
The next nn lines describe the items. Specifically:
- The ii -th line among these starts with two integers, c_{i}ci and d_{i}di ( 1<=d_{i}<c_{i}<=10^{9}1<=di<ci<=109 ), the price of the ii -th good and the discount when using the coupon for the ii -th good, respectively.
- If i>=2i>=2 , this is followed by another integer, x_{i}xi ( 1<=x_{i}<i1<=xi<i ), denoting that the x_{i}xi -th coupon must also be used before this coupon can be used.
输出格式：
Output a single integer on a line by itself, the number of different goods Karen can buy, without exceeding her budget.
输入输出样例
输入样例#1：
6 16
10 9
10 5 1
12 2 1
20 18 3
10 2 3
2 1 5
输出样例#1：
4
输入样例#2：
5 10
3 1
3 1 1
3 1 2
3 1 3
3 1 4
输出样例#2：
5
说明
In the first test case, Karen can purchase the following 44 items:
- Use the first coupon to buy the first item for 10-9=110−9=1 dollar.
- Use the third coupon to buy the third item for 12-2=1012−2=10 dollars.
- Use the fourth coupon to buy the fourth item for 20-18=220−18=2 dollars.
- Buy the sixth item for 22 dollars.
The total cost of these goods is 1515 , 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.
题解
一眼看出树形dp,但是又不知道怎么打,所以考场打了个top序dp后来被证明是错的
知道了是树形dp后怎么定义状态呢?我们看到题目中有一句话
为了使用i张优惠券，凯伦必须也使用第xi张优惠券
很多人可能会这么定义dp数组,(dp[i][j][0])表示第i个节点已经选了j个,本节点不选的最小花费,(dp[i][j][1])就是本节点选的最小花费
可是这样并不能保证本节点选了就一定可以优惠,也就是说是有后效性的
所以我们转换一下思路,(dp[i][j][0])表示第(i)个节点已经选了j个本节点无优惠的最小花费,同理知道(dp[i][j][1])的定义
那么就很好转移了,如下
设(w[i])为(i)物品的原价,(v[i])为优惠后的价格
至于初始值,(dp[u][1][0]=w[u],dp[u][1][1]=v[u],dp[u][0][0]=0)
注意:这道题不能处理出叶子节点的个数,再(dp),会T,要一边dfs一边dp
#include<bits/stdc++.h>
#define Max(a,b) (a)>(b)?(a):(b)
#define Min(a,b) (a)<(b)?(a):(b)
#define in(i) (i=read())
using namespace std;
int read() {
int ans=0,f=1; char i=getchar();
while(i<'0' || i>'9') {if(i=='-') f=-1; i=getchar();}
while(i>='0' && i<='9') {ans=(ans<<1)+(ans<<3)+(i^48); i=getchar();}
return ans*f;
}
int n,s,cnt;
struct node {
int v,w;
}a[5010];
struct edge {
int to,next;
}e[10010];
int head[5010],son[5010];
int dp[5010][5010][2];
void add(int a,int b) {
e[++cnt].to=b;
e[cnt].next=head[a];
head[a]=cnt;
}
void dfs(int u) {
son[u]=1; dp[u][1][0]=a[u].w;
dp[u][1][1]=a[u].v; dp[u][0][0]=0;
for(int i=head[u];i;i=e[i].next) {
int to=e[i].to; dfs(to);
for(int j=son[u];j>=0;--j)//倒序枚举,0/1背包正常枚举方式
for(int k=0;k<=son[to];++k) {
dp[u][j+k][0]=Min(dp[u][j+k][0],dp[u][j][0]+dp[to][k][0]);
dp[u][j+k][1]=Min(dp[u][j+k][1],dp[u][j][1]+dp[to][k][1]);
dp[u][j+k][1]=Min(dp[u][j+k][1],dp[u][j][1]+dp[to][k][0]);
}
son[u]+=son[to];
}
}
int main()
{
//freopen("shopping.in","r",stdin);
//freopen("shopping.out","w",stdout);
in(n); in(s);
memset(dp,0x3f,sizeof(dp));
in(a[1].w); in(a[1].v);
a[1].v=a[1].w-a[1].v;
for(int i=2;i<=n;++i) {
in(a[i].w); in(a[i].v);
a[i].v=a[i].w-a[i].v;
int fa; in(fa); add(fa,i);
} dfs(1);
for(int i=n;i>=0;--i)
if(dp[1][i][0]<=s || dp[1][i][1]<=s)
printf("%d
",i),exit(0);
}