POJ 1837 二维背包 天平平衡问题

Balance

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12362		Accepted: 7745

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

2

Source

题意：已知一个天平 存在n个挂钩   （负数为左挂钩的臂距，正数为右挂钩的臂距）

然后是m个钩码 的重量  

求有多少个放的方法可以使天平 平衡。

因此可以定义一个 状态数组dp[i][j]，意为在挂满前i个钩码时，平衡度为j的挂法的数量。

因此为了不让下标出现负数，做一个处理，使使得数组开为 dp[1~20][0~15000]，则当j=7500时天枰为平衡状态

dp[i][j]+=dp[i-1][j+a[k]*p[i]];

//此题注意变量下标别写混了

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[30],p[30];
int dp[25][15000];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int j=1;j<=m;j++)
            scanf("%d",&p[j]);
        memset(dp,0,sizeof(dp));
        dp[0][7500]=1;
        int Max=-1;
        for(int i=1;i<=m;i++)
        {
            for(int j=15000;j>=0;j--)
            {
                for(int k=1;k<=n;k++)
                {   //判断是否符合范围                            //判断是否前一个状态是否存在 ，不存在就没意义去算（因为为0）
                    if(j+a[k]*p[i]>=0&&j+a[k]*p[i]<=15000&&dp[i-1][j+a[k]*p[i]])
                    dp[i][j]+=dp[i-1][j+a[k]*p[i]];
                  //  Max=max(dp[i][j],Max);
                }
            }
        }
        printf("%d\n",dp[m][7500]);
    }
    return 0;
}

int main()     //此题注意变量下标别写混了
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int j=1;j<=m;j++)
            scanf("%d",&p[j]);

        memset(dp,0,sizeof(dp));
        dp[0][7500]=1;

        for(int i=1;i<=m;i++)
        {                 //20*25*30
            for(int j=0;j<=15000;j++)
            {
                if(dp[i-1][j])
                for(int k=1;k<=n;k++)
                  dp[i][j+a[k]*p[i]]+=dp[i-1][j];
                //printf("%d\n",dp[i][j]);
            }
        }
        printf("%d\n",dp[m][7500]);
    }
    return 0;
}