poj1837(01背包)

Balance

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12867		Accepted: 8071

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：给出c个挂钩在坐标轴上的位置，挂钩能挂砝码，给出g个砝码，挂砝码一定要把所有的砝码挂完，砝码上也可以继续挂砝码，现在求让天平平衡的方法总共有多少种？

思路：

01背包

dp[i][j]表示放置第i个物品后，总质量为j的可能组合的数量。

状态转移方程 dp[i+1][j]=sum{dp[i][j-wt[i]*h[k]](1<=k<=C)}  

初始化 dp[0][0]=1，最后输出 dp[g][0].  

因为dp[i][j]中j可能为负数，在代码中给j加上一个偏移量base。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
const int base=6000;
int h[21],w[21];
int dp[21][base*2];
int main()
{
    int g,c;
    while(~scanf("%d%d",&c,&g))
    {
        memset(dp,0,sizeof(dp));
        for(int i=0; i<c; i++)
            scanf("%d",&h[i]);
        for(int i=0; i<g; i++)
            scanf("%d",&w[i]);
        dp[0][base]=1;
        for(int i=1; i<=g; i++)
            for(int j=-4000+base; j<4000+base; j++)
                for(int k=0; k<c; k++)
                    dp[i][j]+=dp[i-1][j-h[k]*w[i-1]];
        printf("%d\n",dp[g][base]);
    }
    return 0;
}