3312. [USACO13NOV]No Change【狀壓DP】
Description
Farmer John is at the market to purchase supplies for his farm. He has in his pocket K coins (1 <= K <= 16), each with value in the range 1..100,000,000. FJ would like to make a sequence of N purchases (1 <= N <= 100,000), where the ith purchase costs c(i) units of money (1 <= c(i) <= 10,000). As he makes this sequence of purchases, he can periodically stop and pay, with a single coin, for all the purchases made since his last payment (of course, the single coin he uses must be large enough to pay for all of these). Unfortunately, the vendors at the market are completely out of change, so whenever FJ uses a coin that is larger than the amount of money he owes, he sadly receives no changes in return! Please compute the maximum amount of money FJ can end up with after making his N purchases in sequence. Output -1 if it is impossible for FJ to make all of his purchases.
K個硬幣，要買N個物品。
給定買的順序，即按順序必須是一路買過去，當選定買的東西物品序列後，付出錢後，貨主是不會找零錢的。現希望買完所需要的東西後，留下的錢越多越好，如果不能完成購買任務，輸出-1
Input
Line 1: Two integers, K and N.
* Lines 2..1+K: Each line contains the amount of money of one of FJ‘s coins.
* Lines 2+K..1+N+K: These N lines contain the costs of FJ‘s intended purchases.
Output
* Line 1: The maximum amount of money FJ can end up with, or -1 if FJ cannot complete all of his purchases.
Sample Input
3 612
15
10
6
3
3
2
3
7
INPUT DETAILS: FJ has 3 coins of values 12, 15, and 10. He must make purchases in sequence of value 6, 3, 3, 2, 3, and 7.
Sample Output
OUTPUT DETAILS: FJ spends his 10-unit coin on the first two purchases, then the 15-unit coin on the remaining purchases. This leaves him with the 12-unit coin. 首先一看數據範圍狀壓沒跑了 那麽復雜度一定帶一個2^16也就是六萬多 那DP肯定不能和N搞了…… 所以我們就和K搞DP好了 這樣就很容易定義f[i][S]表示當前用了i個硬幣，硬幣使用狀態為S的時候最多買到哪個商品 再預處理pay[i][j]表示硬幣i從j商品開始買能買到哪裏 DP式子就很好想嘍。 一開始沒註意-1WA了一發…… 話說我這算不算面向數據範圍編程
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #define N (100000+10) 5 using namespace std; 6 7 int pay[20][N],a[N],c[N],num[N],f[20][N]; 8 int n,m; 9 10 int main() 11 { 12 scanf("%d%d",&n,&m); 13 for (int i=1; i<=n; ++i) 14 scanf("%d",&a[i]); 15 for (int i=1; i<=m; ++i) 16 scanf("%d",&c[i]); 17 for (int i=1; i<=(1<<n)-1; ++i) 18 { 19 int x=i,cnt=0; 20 while (x){if (x&1) cnt++; x>>=1;} 21 num[i]=cnt; 22 } 23 24 25 int p=0; 26 for (int i=1; i<=n; ++i) 27 { 28 int sum=0,l=1; 29 for (int j=1; j<=m; ++j) 30 { 31 sum-=c[j-1]; 32 while (sum+c[l]<=a[i] && l<=m) 33 sum+=c[l++]; 34 pay[i][j]=l-1; 35 } 36 } 37 38 for (int i=1; i<=n; ++i)//當前硬幣 39 for (int j=0; j<=(1<<n)-1; ++j)//上一個的狀態 40 if (num[j]==i-1) 41 for (int k=1; k<=n; ++k) 42 if ((j|(1<<k-1))!=j) 43 f[i][j|(1<<k-1)]=max(f[i][j|(1<<k-1)],pay[k][f[i-1][j]+1]); 44 45 int ans=-1; 46 for (int i=1; i<=n; ++i) 47 for (int j=1; j<=(1<<n)-1; ++j) 48 if (f[i][j]==m) 49 { 50 int sum=0; 51 for (int k=1; k<=n; ++k) 52 if (!(j&(1<<k-1))) 53 sum+=a[k]; 54 ans=max(ans,sum); 55 } 56 printf("%d",ans); 57 }
3312. [USACO13NOV]No Change【狀壓DP】