BZOJ3312:[USACO]No Change(状压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 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.

Solution

首先一看数据范围状压没跑了

那么复杂度一定带一个2^16也就是六万多

那DP肯定不能和N搞了……

所以我们就和K搞DP好了

这样就很容易定义f[i][S]表示当前用了i个硬币，硬币使用状态为S的时候最多买到哪个商品

再预处理pay[i][j]表示硬币i从j商品开始买能买到哪里

DP式子就很好想喽。

一开始没注意-1WA了一发……

话说我这算不算面向数据范围编程

Code

#include<iostream>
#include<cstring>
#include<cstdio>
#define N (100000+10)
using namespace std;

int pay[20][N],a[N],c[N],num[N],f[20][N];
int n,m;

int main()
{
    scanf("%d%d",&n,&m);
    for (int i=1; i<=n; ++i)
        scanf("%d",&a[i]);
    for (int i=1; i<=m; ++i)
        scanf("%d",&c[i]);
    for (int i=1; i<=(1<<n)-1; ++i)
    {
        int x=i,cnt=0;
        while (x){if (x&1) cnt++; x>>=1;}
        num[i]=cnt;
    }
    
    
    int p=0;
    for (int i=1; i<=n; ++i)
    {
        int sum=0,l=1;
        for (int j=1; j<=m; ++j)
        {
            sum-=c[j-1];
            while (sum+c[l]<=a[i] && l<=m)
                sum+=c[l++];
            pay[i][j]=l-1;
        }
    }
    
    for (int i=1; i<=n; ++i)//当前硬币 
        for (int j=0; j<=(1<<n)-1; ++j)//上一个的状态 
            if (num[j]==i-1)
                for (int k=1; k<=n; ++k)
                    if ((j|(1<<k-1))!=j)
                        f[i][j|(1<<k-1)]=max(f[i][j|(1<<k-1)],pay[k][f[i-1][j]+1]);
    
    int ans=-1;
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=(1<<n)-1; ++j)
            if (f[i][j]==m)
            {
                int sum=0;
                for (int k=1; k<=n; ++k)
                    if (!(j&(1<<k-1)))
                        sum+=a[k];
                ans=max(ans,sum);
            }
    printf("%d",ans);
}