poj 1837 Balance(背包)

题目链接：http://poj.org/problem?id=1837
Balance

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10983		Accepted: 6824

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

题目大意：有一个天平左右两边有C个挂钩，G个钩码。求将全部钩码都挂在钩子上是天平平衡的方法数。这个题目看过去真的是没想法，想想看考完六级，耐心的看完题意，但是能想到的只有用搜索，但是20^20的复杂度超时0.0

想了许久，用最简单的动态规划来写。动态规划的思想就是改变状态的时刻可以从前几状态的推出来。首先要先定义一个平衡度j，j=0表示天平平衡，j>0表示天平右偏，j<0表示天平向左倾。

其次，定义一个状态数组dp[i][j]，表示挂满i个钩码的时候，平衡度为j时的挂法种数。那么每次挂上一个钩码后，对平衡状态的影响因素就是每个钩码的力臂

　力臂=重量 *臂长 = w[i]*c[k]；那么若在挂上第i个砝码之前，天枰的平衡度为j则挂上第i个钩码后，即把前i个钩码全部挂上天枰　　后，天枰的平衡度 j=j+ w[i]*c[k]

　特别注意：最极端的情况是所有物体都挂在最远端，因此平衡度最大值为15*20*25=7500。原则上就应该有dp[ 0..20 ][-7500 .. 　7500 ]。因此做一个处理，使得数组开为 dp[0.. 20][0..15000]。

　详见代码。

 

1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4
5 using namespace std;
6
7 int dp[21][15100];
8
9 int main()
10 {
11     int C,G;
12     int c[21],w[21];
13     while (~scanf("%d%d",&C,&G))
14     {
15         for (int i=1; i<=C; i++)
16         {
17             scanf("%d",&c[i]);
18         }
19         for (int j=1; j<=G; j++)
20         {
21             scanf ("%d",&w[j]);
22         }
23         memset(dp,0,sizeof(dp));
24         dp[0][7500]=1;
25         for (int i=1; i<=G; i++)
26             for (int j=0; j<=15000; j++)
27             {
28                 //cout<<1111<<endl;
29                     for (int k=1; k<=C; k++)
30                         dp[i][j+w[i]*c[k]]+=dp[i-1][j];
31             }
32         printf ("%d\n",dp[G][7500]);
33     }
34     return 0;
35 }