poj3260(多重背包+完全背包)

Description

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, ..., V_N (1 ≤ V_i ≤ 120). Farmer John is carrying C₁ coins of value V₁, C₂ coins of value V₂, ...., and C_N coins of value V_N (0 ≤ C_i ≤ 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).

Input

Line 1: Two space-separated integers: N and T.
Line 2: N space-separated integers, respectively V ₁, V ₂, ..., V_N coins ( V ₁, ... V_N)
Line 3: N space-separated integers, respectively C ₁, C ₂, ..., C_N

Output

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.

Sample Input

3 70
5 25 50
5 2 1

Sample Output

Hint

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.

题意：

John去买东西，东西的价格是T(1 <= T <= 10000)，John所在的地方有n(1 <= n <= 100)种的硬币，面值分别为V1, V2, ...,Vn (1 <= Vi <= 120)。John带了C1枚面值为V1的硬币，C2枚面值为V2的硬币，...，Cn枚面值为Vn的硬币(0 <= Ci <= 10000)。售货员那里每种硬币都有无限多个。问为了支付这个T，John给售货员的硬币数目加上售货员找回的零钱的硬币数目最少是多少。如果无法支付 T，输出-1 。

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思路：

思路： 1.给钱时，硬币数有限制，为多重背包问题。
2.找钱时，硬币数无限制，为完全背包问题。
3.给钱上界为：T+maxValue^2，其中maxValue为最大硬币面值

代码是别人的这里只粘贴出来大家对比这看吧：

代码1(AC 172ms)：

const int T1=14401;
const int T2=T1+14400+2;
const int N=101;
long long sum;
int change[T1],dp[T2];//商店找钱i所需要的硬币数，和付i钱的所要硬币数。
int v[N],c[N];
int bag,n;
int min(int a,int b)
{
    if(a==-1)return b;
    return a>b?b:a;
}
void init(int n)//完全背包，change[]中是找钱的硬币数
{
    memset(change,-1,sizeof(change));
    change[0]=0;
    sum=0;
    for(int i=0;i<n;i++)
    {
       for(int j=v[i];j<T1;j++)
       {
           if(change[j-v[i]]!=-1)
           change[j]=min(change[j],change[j-v[i]]+1);
       }
    }

}
void zeroone(int cost,int val)
{
    __int64 i;
    for( i=sum;i>=cost;i--)
    {
        if(dp[i-cost]!=-1)
            dp[i]=min(dp[i],dp[i-cost]+val);
    }
}     //01背包
int main()
{
    while(cin>>n>>bag)
    {
        for(int i=0;i<n;i++)
            cin>>v[i];
        for(int i=0;i<n;i++)
            cin>>c[i];
        if(bag==0)
        {
            cout<<'0'<<endl;
            continue;
        }
        init(n);
        memset(dp,-1,sizeof(dp));
        dp[0]=0;
        sum=0;
        for(int i=0;i<n;i++)//多重背包
        {
            sum=sum+v[i]*c[i];
            if(sum>=T2)
                sum=T2-1;
            int count1=1;
            while(count1<c[i])
            {
                zeroone(v[i]*count1,count1);
                c[i]-=count1;
                count1<<=1;
            }
            zeroone(v[i]*c[i],c[i]);
        }//利用2进制优化多重背包变为01背包
        int ans=-1;
        for(int i=bag;i<=sum;i++)
        {
            if(dp[i]!=-1&&(i-bag)<T1&&change[i-bag]!=-1)
            {
                ans=min(ans,dp[i]+change[i-bag]);
            }
        }
          cout<<ans<<endl;
    }
}

代码2（AC 94ms）：

//f1[i]表示支付i元所需的最小硬币数
    //f2[i]表示找i元所需的最小硬币数
    //买T元东西所需的最小硬币数为f1[T + i] - f2[i]
    //f1[i] = min(f1[i], f1[i - k] + 1)
    //f2[i] = min(f2[i], f2[i - k] + 1)
    //其中f1[i]多重背包，用二进制拆分求解
    //f2[i]为完全背包
    //其中：给钱上界为：T+maxValue^2，其中maxValue为最大硬币面值。
    //证明：反证法。假设存在一种支付方案，John给的钱超过T+maxValue^2，
    //则售货员找零超过maxValue^2，则找的硬币数目超过maxValue个，将其看作一数列，
    //求前n项和sum(n)，根据鸽巢原理，至少有两 个对maxValue求模的值相等，
    //假设为sum(i)和sum(j),i<j，则i+1...j的硬币面值和为maxValue的倍数，
    //同理，John给的钱中也有 一定数量的硬币面值和为maxValue的倍数，
    //则这两堆硬币可用数量更少的maxValue面值硬币代替，产生更优方案
    const int MAXN = 101;
    const int MAXT = 10000 + 120 * 120 + 1;
    const int INF = 200000000;
    int f1[MAXT];
    int f2[MAXT];
    int v[MAXN];
    int c[MAXN];
    int main() {
    int n, T;
    while(cin >> n >> T) {
    int maxV = 0;
    for(int i = 0; i < n; i++) {
                cin >> v[i];
                maxV = max(maxV, v[i]);
            }
            maxV *= maxV;
    int maxT = T + maxV;
    for(int i = 0; i < n; i++)
                cin >> c[i];
    for(int i = 0; i <= maxT; i++)
                f1[i] = f2[i] = INF;
            f1[0] = f2[0] = 0;
    //完全背包求解
    for(int i = 0; i < n; i++) {
    for(int j = v[i]; j < maxV; j++) {
                    f2[j] = min(f2[j], f2[j - v[i]] + 1);
                }
            }
    //多重背包求解
    for(int i = 0; i < n; i++)
	{
    int k = 1;
    int sum = 0;
    while(sum < c[i])
	{
    for(int j = maxT; j >= v[i] * k; j--)
	{
           f1[j] = min(f1[j], f1[j - v[i] * k] + k);
      }
          sum += k;
    if(sum + k * 2 > c[i])
                        k = c[i] - sum;
    else
                        k *= 2;
     }
     }
    int _min = INF;
    for(int i = T; i <= maxT; i++) {
                _min = min(_min, f1[i] + f2[i - T]);
            }
    if(_min == INF)
                printf("-1\n");
    else
                printf("%d\n", _min);
        }
    return 0;
    }

ps：两种解法一样，先求找零钱的数组，然后找付钱的数组，不同的是数组的初始化和多重背包的写法，其实是一个样子，只是初始化不同，判断条件不同了！

poj3260(多重背包+完全背包)

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