HDU 2844 Coins (多重背包計數 空間換時間)

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CoinsTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8999    Accepted Submission(s): 3623
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

3 101 2 4 2 1 12 51 4 2 10 0

 Sample Output

84

 Source 2009 Multi-University Training Contest 3 - Host by WHU

題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=2844

題目大意：n種數，每個數的值為ai,有ci個，求能組成多少個不大於m的不同的數字
 
題目分析：這題的資料量實在感人，三層迴圈(枚舉種類，枚舉數值，枚舉個數)肯定逾時，要想辦法減去一層迴圈，仔細分析可以發現個數的那層迴圈可以省掉，代價是耗費更多的空間，開一個cnt數組記錄當前第i種使用了多少個，其它兩層迴圈不變，然後就是普通的背包計數問題

#include <cstdio>#include <cstring>int const MAXM = 1e5 + 5;int const MAXN = 1e2 + 5;bool dp[MAXM];int cnt[MAXM];int a[MAXN], c[MAXN];int  main(){    int n, m;    while(scanf("%d %d", &n, &m) != EOF && (n + m))    {        memset(dp, false, sizeof(dp));        for(int i = 1; i <= n; i ++)            scanf("%d", &a[i]);        for(int i = 1; i <= n; i ++)            scanf("%d", &c[i]);        dp[0] = true;        int ans = 0;        for(int i = 1; i <= n; i ++)        {            memset(cnt, 0, sizeof(cnt));            for(int s = a[i]; s <= m; s ++)            {                if(!dp[s] && dp[s - a[i]] && cnt[s - a[i]] < c[i])                {                    dp[s] = true;                    ans ++;                    cnt[s] = cnt[s - a[i]] + 1;                }            }        }        printf("%d\n", ans);    }}

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HDU 2844 Coins (多重背包計數 空間換時間)