POJ 1837Balance(动态规划 好题)

Balance

Time Limit: 1000MS Memory Limit: 30000KTotal 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

给一个天平，然后有许多砝码，问把砝码加上去达到平衡状态的方案数。

测试用例分析：

2 4                   代表有在天平上有两个钩子，砝码有四种

-2 3                  钩子的位置，负数代表在左边，正数代表在右边

3 4 5 8             分别给出四种砝码的重量

什么时候达到平衡呢？就是臂力=臂长*重量，当两边的臂力相等的时候就会平衡了。

题目说的意思是所有的砝码都要用到，我们先分析一下样例。

取第一个砝码 -6 9

取第二个砝码 -14 6 1 21

取第三个砝码 -24 1 -4 21 -9 16 11 36

取第四个砝码 -40 0 0 40

由于第四行太多，没有全部列举，可以看出来，共有两种方案。我们可以使用动态规划的思想，从上层往下层递推。

为了避免负数的出现，我们分析数据g*pos*wei<=20*15*25=7500那么范围就是-7500~7500，那么我们可以使用

dp[25][1500]的背包装这些砝码，状态转移方程为dp[i][j+pos[k]*wei[i]]+=dp[i-1][j];

当然首先得初始化dp[0][7500]=0，标志未放砝码时,7500位置为平衡点。

最后的答案即为dp[g][7500]。

题目地址：Balance

看了几小时，终于看懂了。。哭。

AC代码：

#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;    //一个砝码也不挂        for(i=1; i<=g; i++)   //g个砝码            for(j=0; j<=15000; j++)                if(dp[i-1][j])  //说明可以在往上面挂                    for(k=1; k<=c; k++)   //c个位置                        dp[i][j+wei[i]*pos[k]]+=dp[i-1][j];        cout<<dp[g][7500]<<endl;    }    return 0;}/*2 2-1 12 22 3-1 12 3 5*/