poj 1837(背包weight可能为负数的情况)

Balance

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 9452		Accepted: 5815

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

Source

Romania OI 2002

首先记garr[i]为第i个砝码的重量, carr[i]为第i个挂钩的位置。

用dp[i][j]表示处理[0-i]这些砝码中，力矩和为j的构造方案输，则有：

1、初始化状态为：

dp[0][j] = 1 ,其中j = garr[0] * carr[j]  (j = [0, C])

2、状态转移方程：

dp[i][j] = sum(dp[i][j - garr[i] * carr[k]]) k = [0, C].

注意在具体编程的时候，转移部分的实现代码可以是正向的也可以是逆向的，比如转移方程也可以理解为：

dp[i][j + garr[i] * carr[k]] = sum(dp[i][j]) k = [0, C].

可以发现dp的第二维可能为负数，所以这里展开成dp1和dp2两个数组，分别存储正值和负值。

提交记录：

1、Accepted！

Source Code

Problem: 1837		User: 775700879
Memory: 1524K		Time: 79MS
Language: G++		Result: Accepted
Source Code
#include 
using namespace std;
#define MAXW 7510
int carr[30], garr[30];
int dp1[21][MAXW] = {0};
int dp2[21][MAXW] = {0};
int main() {
    int c, g;
    cin >> c >> g;
    int i, j, k;
    for (i = 0; i < c; i++) cin >> carr[i];
    for (i = 0; i < g; i++) cin >> garr[i];
    for (i = 0; i < c; i++) {
        if (carr[i] * garr[0] >= 0) {
            dp1[0][carr[i] * garr[0]] = 1;
        }
        else {
            dp2[0][-carr[i] * garr[0]] = 1;
        }
    }
    for (i = 1; i < g; i++) {
        for (j = 0; j < MAXW; j++) {
            for (k = 0; k < c; k++) {
                if (dp1[i-1][j] != 0) {
                    if (carr[k] * garr[i] >= 0) {
                        dp1[i][carr[k] * garr[i] + j] += dp1[i-1][j];
                    }
                    else {
                        if (j + carr[k] * garr[i] >= 0) {
                            dp1[i][carr[k]*garr[i] + j] += dp1[i-1][j];
                        }
                        else {
                            dp2[i][-(carr[k]*garr[i]+j)] += dp1[i-1][j];
                        }
                    }
                }
                if (dp2[i-1][j] != 0) {
                    if (carr[k] * garr[i] >= 0) {
                        if (carr[k] * garr[i] - j >= 0) {
                            dp1[i][carr[k]*garr[i] - j] += dp2[i-1][j];
                        }
                        else {
                            dp2[i][j-carr[k]*garr[i]] += dp2[i-1][j];
                        }
                    }
                    else {
                        dp2[i][j-carr[k] * garr[i]] += dp2[i-1][j];
                    }
                }
            }
        }
    }
    cout << dp1[g-1][0] << endl;
}

我事后总觉得 搞两个dp数组有点多余，看了一眼网上的题解，其实这种情况最好直接加上15000，将数组都变成正值，代码量大大缩短：

/*Source Code

Problem: 1837		User: 775700879
Memory: 1580K		Time: 0MS
Language: G++		Result: Accepted
Source Code*/
#include 
using namespace std;
#define MAXW 7510
int carr[30], garr[30];
int dp[21][MAXW*2] = {0};
int main() {
    int c, g;
    cin >> c >> g;
    int i, j, k;
    for (i = 1; i <= c; i++) cin >> carr[i];
    for (i = 1; i <= g; i++) cin >> garr[i];
    dp[0][7500] = 1;
    for (i = 1; i <= g; i++) {
        for (j = 0; j < MAXW*2; j++) {
            if (dp[i-1][j] != 0) 
                for (k = 1; k <= c; k++) {
                    dp[i][j+garr[i]*carr[k]] += dp[i-1][j];
                }
        }
    }
    cout << dp[g][7500] << endl;
}