POJ 1837 Balance (DP)

Balance Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 9627 Accepted: 5939

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

Romania OI 2002

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思路分析：

應該是二維數組，dp[i][j] 表示前i個稱出來j狀態的數量數，j表示重量，小於7500 表示左邊重，反之右邊重。

在此可以用滾動數組。用一個tmp記錄第i個砝碼擴展出來的狀態，然後最後賦給dp
在這裡要說一下，每一次dp都要memset的原因

我自己糾結了好久，因為之前的每一次狀態都是上“一”個砝碼擴展出來的，而且他是要全部都用到。所以比如說

第一個砝碼擴展出來了

-3 3 這兩個狀態，

第二個砝碼得到了轉移 6

那麼 當-3 + 6 之後 -3 這個狀態就不能要了，因為這樣的話只用了一個砝碼。

16MS

#include 
#include 
#include 
#include 

using namespace std;

int dp[17000];
int tmp[17000];
int main()
{
    int c[25];
    int g[25];
    int C,N;
    while(scanf("%d%d",&C,&N)!=EOF)
    {
        memset(dp,0,sizeof dp);

        for(int i=1;i<=C;i++)
        scanf("%d",&c[i]);
        for(int i=1;i<=N;i++)
        scanf("%d",&g[i]);
        dp[7500]=1;

        for(int i=1;i<=N;i++)
        {
            memset(tmp,0,sizeof(tmp));
            for(int k=0;k<=15000;k++)
            if(dp[k])
            for(int j=1;j<=C;j++)
            {
                int v=g[i]*c[j];
                tmp[k+v]+=dp[k];
            }
            memset(dp,0,sizeof dp);
            for(int j=0;j<=15000;j++)
            if(tmp[j])dp[j]+=tmp[j];
        }
        printf("%d\n",tmp[7500]);
    }
    return 0;
}