Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3424 Accepted Submission(s): 1345
Problem Description Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch. You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.Input The input contains several test cases. The first line of each test case contains two integers n(1 ≤ n ≤ 100),m(m ≤ 100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1 ≤ Ai ≤ 100000,1 ≤ Ci ≤ 1000). The last test case is followed by two zeros.
Output For each test case output the answer on a single line.
Sample Input 3 10 1 2 4 2 1 1 2 5 1 4 2 1 0 0
Sample Output 8 4
Source 2009 Multi-University Training Contest 3 - Host by WHU
Recommend gaojie 分析：多重背包问题.有n个物品（硬币）,每个物品有价值value[i],每个物品数量限制为num[i]. 这道题要求的是用所有的硬币在1到m的范围内最多能够组合成多少种价值.每种硬币有数量的限制. 对于每种硬币而言: if 价值×数量>=m then 取这种硬币的次数相当于无限制,可以考虑成完全背包 else then 考虑成0-1背包(二进制优化),就是把这种硬币的value和num组合出0-1背包可能出现的状态(可以去看背包九讲)
(对于num,类似于编码.当2^n<=num/2时:k=2^n(n=0,1,2,……)表示状态,对应下来就是二进制的某一位数是1,然后还有一个状态就是k>num/2的时候啦,num+1-k,这样下来就可以用k来组合枚举出从1->num的所有可能了.然后对于k，单位价值和大小都乘上k之后就变成了一个01背包).
对于多重背包问题可以这样解: MultiplePack(cost,weight,amount) { if(cost*weight>=V) CompletePack(cost,weight) else { for(int i=1;i<amount;i<<1) { ZeroOnePack(i*cost,i*weight); amount-=i; } ZeroOnePack(amount*cost,amount*weight) } }
这道题的cost和weight是同一种量.
#include<iostream> using namespace std; int dp[100010]; int a[110],c[110]; int n,m; void ZeroOnePack(int c,int w)//01背包 { for(int i=m;i>=c;i--) dp[i]=dp[i]|dp[i-c]; /*等价于dp[i]=Max(dp[i],dp[i-c]);*/; } void CompletePack(int c,int w)//完全背包 { for(int i=c;i<=m;i++) dp[i]=dp[i]|dp[i-c]; } void MutiplePack(int c,int w)//多重背包 { int temp; if(c*w>=m)//如果物品的总花费大于背包容量就利用完全背包解决 CompletePack(c,w); else//否则利用01背包解决(二进制思想优化) { temp=w; for(int i=1;i<=temp;i<<2) { ZeroOnePack(i*c,i*w); temp-=i; } ZeroOnePack(temp*c,temp*w); } } int main() { int i; while(scanf("%d%d",&n,&m),n||m) { memset(dp,0,sizeof(dp)); for(i=1;i<=n;i++) scanf("%d",&a[i]); for(i=1;i<=n;i++) scanf("%d",&c[i]); dp[0]=1; for(i=1;i<=n;i++) MutiplePack(a[i],c[i]); int ans=0; for(i=1;i<=m;i++) if(dp[i]) ans++; printf("%d\n",ans); } return 0; }
转载于:https://www.cnblogs.com/Su-Blog/archive/2012/08/19/2646558.html
