POJ 1742 coins

題意：給出幾種面值的錢幣和對應的個數，看能否湊出1-m中的各個面值。

分析：顯然，多重揹包問題。要求全部裝滿。而且對1-m遍歷並計數，毫無壓力~

上代碼：

#include <iostream>
using namespace std;

int  MIN_INT = (~(unsigned(-1)>>1));
int F[100001];
int A[101];
int C[101];

int max(int a, int b)
{
	return a>b?a:b;
}
void ZeroOnePack(int cost, int weight, int V)
{
	for (int v=V; v>=cost; --v)
		F[v] = max(F[v], F[v-cost]+weight);
}

void CompletePack(int cost, int weight, int V)
{
	for (int v=cost; v<=V; ++v)
		F[v] = max(F[v], F[v-cost]+weight);
}

void MultiPack(int cost, int weight, int V, int amount)
{
	if (cost*amount>=V) {
		CompletePack(cost, weight, V);
		return;
	}
	int k = 1;
	while (k<amount) {
		ZeroOnePack(cost*k, weight*k, V);
		amount -= k;
		k *= 2;
	}
	ZeroOnePack(cost*amount, weight*amount, V);
}

int main(int argc, char **argv)
{
	int n, m, count;
	while ((cin>>n>>m) && (m+n)) {
		for (int i=1; i<=n; ++i)
			cin>>A[i];
		for (int i=1; i<=n; ++i)
			cin>>C[i];
		count = 0;
		for (int V=1; V<=m; ++V) {
			F[0] = 0;
			for (int i=1; i<=m; ++i)
				F[i] = MIN_INT;
			for (int i=1; i<=n; ++i)
				MultiPack(A[i], A[i], V, C[i]);

			if (F[V]==V)
				++count;
		}
		cout<<count<<endl;
	}
	system("pause");
	return 0;
}

好吧~POJ的系統說TLE了~悲劇的一米啊~看了discuss說，這題沒那麼簡單O(V*Σlog n[i])是過不了的~好吧，優化優化~

 

再分析：

其實上面的程序求出了F[V]的值，這個理論上是不需要的，我們關心F[V]是否等於V而已。這裏我們進行優化：

將int F[100001]修改爲bool F[100001]，存儲F[V]是否等於V

ZeroOnePack和CompletePack可以修改爲

void ZeroOnePack(int cost, int weight, int V)

{

      for (int v=V; v>=cost; --v)

            F[v] |= F[v-cost];

}

 

void CompletePack(int cost, int weight, int V)

{

      for (int v=cost; v<=V; ++v)

            F[v] |= F[v-cost];

}

F[v] |= F[v-cost] 直接或運算用0,1表示此價格是否出現過

最後上代碼：

#include <iostream>

using namespace std;

bool F[100001];
int A[101];
int C[101];

int max(int a, int b)
{
	return a>b?a:b;
}
void ZeroOnePack(int cost, int weight, int V)
{
	for (int v=V; v>=cost; --v)
		F[v] |= F[v-cost];
}

void CompletePack(int cost, int weight, int V)
{
	for (int v=cost; v<=V; ++v)
		F[v] |= F[v-cost];
}

void MultiPack(int cost, int weight, int V, int amount)
{
	if (cost*amount>=V) {
		CompletePack(cost, weight, V);
		return;
	}
	int k = 1;
	while (k<amount) {
		ZeroOnePack(cost*k, weight*k, V);
		amount -= k;
		k *= 2;
	}
	ZeroOnePack(cost*amount, weight*amount, V);
}

int main(int argc, char **argv)
{
	int n, m;
	while ((cin>>n>>m) && (m+n)) {
		for (int i=1; i<=n; ++i)
			cin>>A[i];
		for (int i=1; i<=n; ++i)
			cin>>C[i];
		
		F[0] = 1;
		for (int i=1; i<=m; ++i)
			F[i] = 0;
		for (int i=1; i<=n; ++i)
			MultiPack(A[i], A[i], m, C[i]);
		
		int ans=0;
		for(int i=1;i<=m;++i)
			ans+=F[i];
		cout<<ans<<endl;
	}
	system("pause");
	return 0;
}
2329MS險過~