Codeforces 815 C Karen and Supermarket
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
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
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.
Example
Input6 16Output
10 9
10 5 1
12 2 1
20 18 3
10 2 3
2 1 5
4Input
5 10Output
3 1
3 1 1
3 1 2
3 1 3
3 1 4
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.
發現依賴關系可以構成一個樹結構,然後就是一個樹上DP了2333
#include<bits/stdc++.h> #define ll long long #define maxn 5005 #define pb push_back using namespace std; vector<int> g[maxn]; int h[maxn][maxn],siz[maxn]; //用優惠券的 int f[maxn][maxn],fa; //不用優惠券的 int n,m,a[maxn],b[maxn],k; inline int min(const int x,const int y){ return x>y?y:x; } void dfs(int x){ int to; siz[x]=1; h[x][1]=b[x]; f[x][0]=0,f[x][1]=a[x]; for(int i=g[x].size()-1;i>=0;i--){ to=g[x][i]; dfs(to); for(int j=siz[x];j>=0;j--){ // if(f[x][j]>k&&h[x][j]>k) break; for(int u=1;u<=siz[to];u++){ if(f[to][u]>k&&h[to][u]>k) break; if(f[to][u]<=k){ h[x][j+u]=min(h[x][j+u],f[to][u]+h[x][j]); f[x][j+u]=min(f[x][j+u],f[to][u]+f[x][j]); } if(h[to][u]<=k) h[x][j+u]=min(h[x][j+u],h[to][u]+h[x][j]); } } siz[x]+=siz[to]; } } int main(){ memset(h,0x3f,sizeof(h)); memset(f,0x3f,sizeof(f)); scanf("%d%d",&n,&k); for(int i=1;i<=n;i++){ scanf("%d%d",a+i,b+i); b[i]=a[i]-b[i]; if(i>1){ scanf("%d",&fa); g[fa].pb(i); } } dfs(1); for(int i=n;i>=0;i--) if(h[1][i]<=k||f[1][i]<=k){ printf("%d\n",i); return 0; } return 0; }
Codeforces 815 C Karen and Supermarket