P3092 [USACO13NOV]没有找零No Change

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

输入输出格式

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.
* 输入输出样例

输入样例#1：
4 0
0 0 1 1
输出样例#1：
512
输入样例#2：
5 0
1 0 1 1 1
输出样例#2：
5120
题解：状态压缩DP+二分优化
直接看代码好了

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define il inline
#define rg register
#define ll long long
#define N 20
#define M 100010
#define inf 2147483640
using namespace std;
int k,n;
ll coin[N],cost[M],sum[M],dp[M],ans=-1;
il int read()
{
    rg int data=0;rg char c=getchar();
    while(c<'0'||c>'9')c=getchar();
    while(c>='0'&&c<='9')data=(data<<3)+(data<<1)+c-'0',c=getchar();
    return data;
}
int find(rg int lim){
    rg int le=1,ri=n,pos=0;
    while(le<=ri){
        int mid=((le+ri)>>1);
        if(sum[mid]<=lim)le=mid+1,pos=mid;
        else ri=mid-1;
    }
    return pos;
}
int main()
{
    freopen("s.in","r",stdin);
    k=read(),n=read();
    for(rg int i=0;i<k;++i)//请注意这里的for循环
        coin[i]=read();
    for(rg int i=1;i<=n;++i)
        cost[i]=read(),sum[i]=sum[i-1]+cost[i];//预处理

    for(rg int p=0;p<(1<<k);++p){//枚举所有状态
        for(rg int i=0;i<k;++i){
            if(!(p&(1<<i)))continue;//如果当前货币i没有被选则跳过
            ll res=dp[p^(1<<i)];//p^(1<<j)表示状态pi去掉ci货币时得到的新状态
            //那么res表示新状态可以到达的最末商品
            ll pos=find(sum[res]+coin[i]);
            //我们期望查找到的是从商品1到res的价值并加上货币ci价值的总和pos
            //二分查找到最后一个小于等于的商品下标
            dp[p]=max(dp[p],pos);
            //取最大值
        }
    }
    for(rg int p=0;p<(1<<k);++p){
        if(dp[p]==n){
            ll cnt=0;
            for(rg int j=0;j<k;++j){
                if(!(p&(1<<j)))cnt+=coin[j];
                //如果当前位置没有货币，统计答案
            }
            ans=max(ans,cnt);
        }
    }
    cout<<ans;
    return 0;
}

总结：被for循环坑了一波，最近状态压缩DP,做到快吐