POJ 1837 Balance（分组背包）

最新推荐文章于 2020-07-05 15:21:19 发布

baymax520

最新推荐文章于 2020-07-05 15:21:19 发布

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分类专栏：背包dp 文章标签： dp

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本文探讨了一个复杂的天平平衡问题，通过分析特定条件下的挂钩位置和砝码重量，提出了一种有效的解决方案。利用动态规划算法，计算在限定条件下天平能够达到平衡的所有可能方式的数量。

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Balance

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16611		Accepted: 10433

Description

Gigel has a strange "balance" and he wants to poise it. Actually, the device is different from any other ordinary balance.
It orders two arms of negligible weight and each arm's length is 15. Some hooks are attached to these arms and Gigel wants to hang up some weights from his collection of G weights (1 <= G <= 20) knowing that these weights have distinct values in the range 1..25. Gigel may droop any weight of any hook but he is forced to use all the weights.
Finally, Gigel managed to balance the device using the experience he gained at the National Olympiad in Informatics. Now he would like to know in how many ways the device can be balanced.

Knowing the repartition of the hooks and the set of the weights write a program that calculates the number of possibilities to balance the device.
It is guaranteed that will exist at least one solution for each test case at the evaluation.

Input

The input has the following structure:
• the first line contains the number C (2 <= C <= 20) and the number G (2 <= G <= 20);
• the next line contains C integer numbers (these numbers are also distinct and sorted in ascending order) in the range -15..15 representing the repartition of the hooks; each number represents the position relative to the center of the balance on the X axis (when no weights are attached the device is balanced and lined up to the X axis; the absolute value of the distances represents the distance between the hook and the balance center and the sign of the numbers determines the arm of the balance to which the hook is attached: '-' for the left arm and '+' for the right arm);
• on the next line there are G natural, distinct and sorted in ascending order numbers in the range 1..25 representing the weights' values.

Output

The output contains the number M representing the number of possibilities to poise the balance.

Sample Input

2 4	
-2 3 
3 4 5 8

Sample Output

Source

Romania OI 2002

题意：有一个天平，天平左右臂长最多15，给c个数字代表挂钩的位置，负数表示在左臂，正数在右臂。在给g个砝码，要求使用所有砝码让天平平衡的方法数。

思路：考虑极限情况，最多个数的最重砝码挂在最边上，平衡度为15*20*25=7500，我们设平衡度v=0的时候为天平平衡状态，左倾v<0，右倾v>0，那么左倾的极端情况为-7500，为了避免数组出现负数，我们将整个坐标右移，设定7500为平衡点。然后每一个砝码为一组物品，挂钩数为每组的物品数，因为每一个砝码只能挂在一个挂钩上面，符合普通分组背包一组只能选一个物品的规则。dp[i][j]表示前i组物品平衡度为j的方案数，转移方程就为dp[i][j]+=dp[i-1][j-c[k]*v[i]]。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int dp[25][15010],c,g,cost[25],value[25];
int main()
{
    scanf("%d%d",&c,&g);
    for(int i=1;i<=c;i++)
        scanf("%d",&cost[i]);
    for(int i=1;i<=g;i++)
        scanf("%d",&value[i]);
    dp[0][7500]=1;
    for(int i=1;i<=g;i++)
    {
        for(int j=15000;j>=0;j--)
        {
            for(int k=1;k<=c;k++)
            {
                if(j-cost[k]*value[i]>=0)
                    dp[i][j]+=dp[i-1][j-cost[k]*value[i]];
            }
        }
    }
    printf("%d\n",dp[g][7500]);
    return 0;
}