bzoj3312 [Usaco2013 Nov]No Change

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个硬币，要买N个物品。
给定买的顺序，即按顺序必须是一路买过去，当选定买的东西物品序列后，付出钱后，货主是不会找零钱的。现希望买完所需要的东西后，留下的钱越多越好，如果不能完成购买任务，输出-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
4000
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 6

12

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

12

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.

k<=16明显是要状态压缩的样啊！！！

f[i]代表硬币的使用状态为i的时候最多能买到第几个商品。之后的转移就很简单了。

代码如下(有点慢):

{$inline on}
var
f:array[0..131072]of longint;
a:array[1..16]of longint;
sum:array[0..100000]of longint;
summ,ans,n,m,i,j,k:longint;
function max(a,b:longint):longint;inline;
begin
if a>b then exit(a) else exit(B);
end;

function find(st,y,x:longint):longint;inline;
var
mid,t,w:longint;
begin
t:=st; w:=n;
while t<=w do
begin
mid:=(t+w) div 2;
if sum[mid]-y>x then
w:=mid-1
else
t:=mid+1;
end;
exit(t-1);
end;

begin
readln(k,n);
for i:=1 to k do
readln(a[i]);
for i:=1 to n do
begin
readln(sum[i]);
sum[i]:=sum[i-1]+sum[i];
end;
for j:=0 to 1<<(k+1)-1 do
for i:=1 to k do
if j and (1<<i)=0 then
f[j or (1<<i)]:=max(f[j or (1<<i)],find(f[j],sum[f[j]],a[i]));
ans:=-1;
for i:=0 to 1<<(k+1)-1 do
if f[i]=n then
begin
summ:=0;
for j:=1 to k do
if i and (1<<j)=0 then
inc(summ,a[j]);
ans:=max(ans,summ);
end;
writeln(ans);
end.