POJ 1276 Cash Machine 多重背包

Cash Machine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 31470   Accepted: 11307 Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the machine has a supply of nk bills. For example,  N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10  means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each.  Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine.  Notes:  @ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc.  Input The program input is from standard input. Each data set in the input stands for a particular transaction and has the format:  cash N n1 D1 n2 D2 ... nN DN  where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers in the input. The input data are correct.  Output For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.  Sample Input 735 3 4 125 6 5 3 350 633 4 500 30 6 100 1 5 0 1 735 0 0 3 10 100 10 50 10 10 Sample Output 735 630 0 0 Hint The first data set designates a transaction where the amount of cash requested is @735. The machine contains 3 bill denominations: 4 bills of @125, 6 bills of @5, and 3 bills of @350. The machine can deliver the exact amount of requested cash.  In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is @630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered cash.  In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is @0 and, therefore, the machine delivers no cash. Source Southeastern Europe 2002

点击打开题目链接：POJ 1276

题意就不说了，n 个物品，每个物品有个数，价值，选一些物品使其在不超过给定价值的前提下获得最大价值

dp[ i ]：标记选一些物品能否组成价值 i，1 表示可以，0 表示不可以

我看出来用完全背包做，然并卵。我没有做出来，搜过题解后将其时间优化减少了一半，你还可以再优化吗？

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int MAXN = 10 + 5;
const int MAXM = 1e5 + 10;
int dp[MAXM];
int num[MAXN], wei[MAXN];

int main()
{
    //freopen("in.txt", "r", stdin);
    int sum, n;
    while (~scanf("%d%d", &sum, &n))
    {
        memset(dp, 0, sizeof(dp));

        for (int i = 1; i <= n; i++)
        {
            scanf("%d%d", &num[i], &wei[i]);
        }
        dp[0] = 1;          //初始化 0 会出现
        int Max = 0, temp = 0;
        for (int i = 1; i <= n; i++)        //物品种类
        {
            for (int j = Max; j >= 0; j--)  //从当前最大值开始
            {
                if (dp[j])                  //如果这个值已经出现过
                    for (int k = 1; k <= num[i]; k++)//那么我们再加上k个第i个物品得到的值
                    {                                   //也一定会出现
                        temp = j + k * wei[i];
                        if (temp > sum) break;      //超了
                        if (dp[temp]) continue;     //已经出现过，直接判断下一个（节省时间）
                        dp[temp] = 1;
                        Max = max(Max, temp);
                    }
            }
        }
        printf("%d\n", Max);
    }
    return 0;
}