POJ 1837Balance(動態規劃 好題)

果7發表於2014-01-19
動態規劃
Balance
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 9615   Accepted: 5933

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


給一個天平，然後有許多砝碼，問把砝碼加上去達到平衡狀態的方案數。

測試用例分析：

2 4                   代表有在天平上有兩個鉤子，砝碼有四種

-2 3                  鉤子的位置，負數代表在左邊，正數代表在右邊

3 4 5 8             分別給出四種砝碼的重量

什麼時候達到平衡呢？就是臂力=臂長*重量，當兩邊的臂力相等的時候就會平衡了。


題目說的意思是所有的砝碼都要用到，我們先分析一下樣例。

取第一個砝碼 -6 9

取第二個砝碼 -14 6 1 21

取第三個砝碼 -24 1 -4 21 -9 16 11 36

取第四個砝碼 -40 0 0 40


由於第四行太多，沒有全部列舉，可以看出來，共有兩種方案。我們可以使用動態規劃的思想，從上層往下層遞推。

為了避免負數的出現，我們分析資料g*pos*wei<=20*15*25=7500那麼範圍就是-7500~7500，那麼我們可以使用

dp[25][1500]的揹包裝這些砝碼，狀態轉移方程為dp[i][j+pos[k]*wei[i]]+=dp[i-1][j];

當然首先得初始化dp[0][7500]=0，標誌未放砝碼時,7500位置為平衡點。

最後的答案即為dp[g][7500]。

題目地址：Balance

看了幾小時，終於看懂了。。哭。

AC程式碼：

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

int dp[25][15001];
int pos[25],wei[25];

int main()
{
    int c,g;
    int i,j,k;

    while(cin>>c>>g)
    {
        for(i=1; i<=c; i++)
            cin>>pos[i];
        for(i=1; i<=g; i++)
            cin>>wei[i];

        memset(dp,0,sizeof(dp));
        dp[0][7500]=1;    //一個砝碼也不掛

        for(i=1; i<=g; i++)   //g個砝碼
            for(j=0; j<=15000; j++)
                if(dp[i-1][j])  //說明可以在往上面掛
                    for(k=1; k<=c; k++)   //c個位置
                        dp[i][j+wei[i]*pos[k]]+=dp[i-1][j];

        cout<<dp[g][7500]<<endl;
    }
    return 0;
}

/*
2 2
-1 1
2 2
2 3
-1 1
2 3 5
*/



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