[USACO06DEC]The Fewest Coins G

题目描述

Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he uses to pay plus the number of coins he receives in change is minimized. Help him to determine what this minimum number is.

FJ wants to buy T (1 ≤ T ≤ 10,000) cents of supplies. The currency system has N (1 ≤ N ≤ 100) different coins, with values V1, V2, …, VN (1 ≤ Vi ≤ 120). Farmer John is carrying C1 coins of value V1, C2 coins of value V2, …, and CN coins of value VN (0 ≤ Ci ≤ 10,000). The shopkeeper has an unlimited supply of all the coins, and always makes change in the most efficient manner (although Farmer John must be sure to pay in a way that makes it possible to make the correct change).

农夫John想到镇上买些补给。为了高效地完成任务，他想使硬币的转手次数最少。即使他交付的硬币数与找零得到的的硬币数最少。

John想要买价值为T的东西。有N(1<=n<=100)种货币参与流通，面值分别为V1,V2…Vn (1<=Vi<=120)。John有Ci个面值为Vi的硬币(0<=Ci<=10000)。

我们假设店主有无限多的硬币，并总按最优方案找零。注意无解输出-1。

输入格式

Line 1: Two space-separated integers: N and T.

Line 2: N space-separated integers, respectively V1, V2, …, VN coins (V1, …VN)

Line 3: N space-separated integers, respectively C1, C2, …, CN

输出格式

Line 1: A line containing a single integer, the minimum number of coins involved in a payment and change-making. If it is impossible for Farmer John to pay and receive exact change, output -1.

样例 #1

样例输入 #1

3 70
5 25 50
5 2 1

样例输出 #1

提示

Farmer John pays 75 cents using a 50 cents and a 25 cents coin, and receives a 5 cents coin in change, for a total of 3 coins used in the transaction.

明显购买是一个多重背包（二进制优化及以上），找零是一个完全背包，唯一的问题是确定购买的最大 MMM 。

题解中说 M=max(v[i]2)M = max(v[i]^2)M=max(v[i]2) 我不会证明，也没看懂，不过， M=∑v[i]M = \sum v[i]M=∑v[i] 是明显成立的，然后跑一个枚举就行。

/*
A: 10min
B: 20min
C: 30min
D: 40min
*/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <assert.h>
#include <sstream>
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define mem(f, x) memset(f,x,sizeof(f))
#define fo(i,a,n) for(int i=(a);i<=(n);++i)
#define fo_(i,a,n) for(int i=(a);i<(n);++i)
#define debug(x) cout<<#x<<":"<<x<<endl;
#define endl '\n'
using namespace std;
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")template<typename T>
ostream& operator<<(ostream& os,const vector<T>&v){for(int i=0,j=0;i<v.size();i++,j++)if(j>=5){j=0;puts("");}else os<<v[i]<<" ";return os;}
template<typename T>
ostream& operator<<(ostream& os,const set<T>&v){for(auto c:v)os<<c<<" ";return os;}
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const map<T1,T2>&v){for(auto c:v)os<<c.first<<" "<<c.second<<endl;return os;}
template<typename T>inline void rd(T &a) {char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}typedef pair<long long ,long long >PII;
typedef pair<long,long>PLL;typedef long long ll;
typedef unsigned long long ull;
const int N=2e5+10,M=1e9+7;ll n,m;
struct good{int v,w;
};
vector<good>g;
int V[110],C[110],f[21000],dp[21000],mx;
void solve(){cin>>n>>m;memset(f,0x3f,sizeof f); // f[j]表示恰好购买j价值物品所需要的最小支付硬币数量 , 多重背包memset(dp,0x3f,sizeof dp); // 完全背包f[0] = 0;dp[0] = 0;fo(i,1,n)cin>>V[i],mx = max(mx,V[i]);mx *= mx;fo(i,1,n)cin>>C[i];fo(i,1,n){int s = C[i];for(int k=1;k<=C[i];k<<=1){g.pb({k*V[i],k});s-=k;}if(s>0){g.pb({s*V[i],s});}}for(auto c:g){for(int j=mx;j>=c.v;j--){// J 是钱f[j] = min(f[j],f[j-c.v]+c.w);}for(int j=c.v;j<=mx;j++){dp[j] = min(dp[j],dp[j-c.v]+c.w);}}if(f[m] == 0x3f3f3f3f){puts("-1");}else{int ans = 0x3f3f3f3f;for(int i=m;i<=m+mx;i++){debug(i);ans = min(ans,f[i]+dp[i-m]);}if(ans == 0x3f3f3f3f)ans=-1;cout<<ans;}
}
int main(){solve();return 0;
}