poj-1837-Balance

Balance
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 14387 Accepted: 9059
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

题意：c个挂钩，g个钩码，要让天平左右保持平衡，输出钩码全挂上平衡的方法总数。

思路：
每向天平上挂一个钩码即可得到一种状态，这种状态可由前一个状态获得
用dp[i][j]表示前i个钩码状态j的方法数，j<0表示天平左边重，j>0则相反，j=0平衡。
避免负数，将j=7500设置为平衡状态。
7500=15*20*25（距离范围1~15，钩码数最多20，重量最大25）
则数组开dp[21][15000] 。

设dp[i-1][j]=num，那么dp[i][j]有两种策略：即挂或不挂，不难得到：dp[i][j]+=dp[i-1][j-weight[i]*pos[k]]。前提是j>=weight[i]*pos[k]

优化：
假设 dp[i-1][j] 的值已知，设dp[i-1][j]=num（即已知把前i-1个钩码全部挂上天枰后得到状态j的方法有num次）
那么dp[i][ j+ weight[i]*pos[k] ] = dp[i-1][j] = num（即以此为前提，在第k个钩子挂上第i个钩码后，得到状态j+ weight[i]*pos[k]的方法也为num次）
利用递归思想，不难得出 状态方程dp[i][ j+ weight[i]*pos[k] ]+= dp[i-1][j]

代码：

#include <stdio.h>#include <string.h>int dp[21][15000];//挂上前i个钩码后，达到j状态的方法数  int main(){    int i,j,k,c,g;    int pos[21],weight[21];    while(~scanf("%d%d",&c,&g))    {          for(i=1;i<=c;i++)            scanf("%d",&pos[i]);        for(i=1;i<=g;i++)            scanf("%d",&weight[i]);        memset(dp,0,sizeof(dp));  //达到每个状态的方法数初始化为0          dp[0][7500]=1; //7500为天枰达到平衡状态时的平衡度                         //放入前0个物品后，天枰达到平衡状态7500的方法有1个，就是不挂钩码        for(i=1;i<=g;i++)            for(j=0;j<=15000;j++) //          {//              for(k=1;k<=c;k++){//                  if(j>=pos[k]*weight[i]){//                      dp[i][j]+=dp[i-1][j-pos[k]*weight[i]];//容易得到 dp[i][j]=dp[i][j]+dp[i-1][j-pos[k]*weight[i]]//                  }//              }//            }                if(dp[i-1][j]){//优化，当放入i-1个物品时状态j已经出现且被统计过方法数，则直接使用统计结果                      //否则忽略当前状态j                      for(k=1;k<=c;k++)                          dp[i][j+pos[k]*weight[i]]+=dp[i-1][j];                }          printf("%d\n",dp[g][7500]);    }    return 0; }