P3092 [USACO13NOV]No Change G

[USACO13NOV]No Change G

题目描述

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.

约翰到商场购物，他的钱包里有K(1 <= K <= 16)个硬币，面值的范围是1..100,000,000。

约翰想按顺序买 N个物品(1 <= N <= 100,000)，第i个物品需要花费c(i)块钱，(1 <= c(i) <= 10,000)。

在依次进行的购买N个物品的过程中，约翰可以随时停下来付款，每次付款只用一个硬币，支付购买的内容是从上一次支付后开始到现在的这些所有物品（前提是该硬币足以支付这些物品的费用）。不幸的是，商场的收银机坏了，如果约翰支付的硬币面值大于所需的费用，他不会得到任何找零。

请计算出在购买完N个物品后，约翰最多剩下多少钱。如果无法完成购买，输出-1

输入格式

* 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.

输出格式

* Line 1: The maximum amount of money FJ can end up with, or -1 if FJ cannot complete all of his purchases.

样例 #1

样例输入 #1

3 6 
12 
15 
10 
6 
3 
3 
2 
3 
7

样例输出 #1

12

提示

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.

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.

思路

数据范围16好像暴力搜索是16！感觉有点不稳
所以我们思考其他方式
我们思考可以好像要是状态dp的话转移很方便
我们设dp[i]为 状态i时能买到哪
我们只需要从小到大枚举 每一次从减去最后的1转移
转移也只是将当前 而这样转移的时间复杂度是O（1e51e5)的
我们考虑优化转移 我们可以额外维护一个f[i]表示 i状态时我们买物品花了多少钱
然后我们可以维护物品的前缀和 二分查找即可
代码里是直接按位枚举的 O（(1<<16)16logN）
当然可以直接lowbit 就会优化掉16 O（(1<<16)logN）

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+10;
const int M = 998244353;
const int mod = 1e9+7;
#define int long long
#define endl '\n'
#define all(x) (x).begin(),(x).end()
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define _ 0
#define pi acos(-1)
#define INF 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
int dp[1<<16],f[1<<16];//dp[i]表示i状态下的能买到哪 f[i]表示i状态的买到哪的cost sum
void solve() {
    int n,m;cin>>n>>m;
    vector<int>a(n+1),c(m+1),sum(n+1),cum(m+1);
    for(int i=1;i<=n;i++)cin>>a[i],sum[i]+=a[i]+sum[i-1];
    for(int i=1;i<=m;i++)cin>>c[i],cum[i]+=c[i]+cum[i-1];
    int ans=-1;
    for(int i=0;i<(1<<n);i++)
        for(int j=0;j<n;j++)if(i>>j&1) {
                int t = i - (1 << j);
                int now = a[j+1] + f[t];
                int k = upper_bound(all(cum), now) - cum.begin();
                k--;
                dp[i]=max(dp[i],k);
                f[i]=max(f[i],cum[k]);
                if (k == m) {
                    int res = 0;
                    for (int r = 0; r < n; r++)if (i >> r & 1)res += a[r+1];
                    ans = max(ans, sum[n]-res);
                }
            }
    cout<<ans<<endl;
}
signed main(){
    fast
    int T;T=1;
    while(T--) {
        solve();
    }
    return ~~(0^_^0);
}