POJ 1837 Balance（動態規劃之背包問題）

POJ 1837 Balance（動態規劃之背包問題）

Balance Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 11436   Accepted: 7130

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

題意：一個天平上有C個掛鉤，第i個掛鉤的位置為C[i]，C[i] < 0表示該掛鉤在原點的左邊，C[i] > 0表示該掛鉤在原點的右邊；然後給出G個鉤碼的重量，問有多少種掛法使得天平保持平衡。

 

分析：當天平平衡時，每向天平上掛一個鉤碼，天平的狀態就會改變，而這個狀態可以由若干前一狀態獲得。

首先定義一個平衡度j的概念

當平衡度j=0時，說明天枰達到平衡，j>0，說明天枰傾向右邊（x軸右半軸），j<0則相反

那麼此時可以把平衡度j看做衡量當前天枰狀態的一個值

因此可以定義一個 狀態數組dp[i][j]，意為在掛滿前i個鉤碼時，平衡度為j的掛法的數量。

由於距離L[i]的范圍是-15~15，鉤碼重量的范圍是w[i]是1~25，鉤碼數量最大是20

因此最極端的平衡度是所有物體都掛在最遠端，因此平衡度最大值為j=15*20*25=7500。原則上就應該有dp[ 1~20 ][-7500 ~ 7500 ]。

因此為了不讓下標出現負數，做一個處理，使得數組開為 dp[1~20][0~15000]，令7500對應0，則當j=7500時天枰為平衡狀態。

 

/*
    dp[i][j]表示用了前i個鉤碼，天平兩端的差值（右端 - 左端）為j時的方案數。
    極端情況為所有鉤碼全部掛在左端的最左邊位置，根據題目數據可知此時差值最大為
    0 - 15 * 25 * 20 = -7500，所以要加上一個偏移量，用7500對應0，假設鉤碼數量為G，
    則最終答案為dp[G][7500]。
    轉移方程：dp[i][j + w[i] * l[k]] = Sigma(dp[i-1][j])。
    k為第k個掛鉤位置
*/
#include 
#include 
#include 
using namespace std;

int dp[21][15005];
int l[21], w[21];

int main() {
    int C, G;
    while(~scanf("%d%d", &C, &G)) {
        for(int i = 1; i <= C; i++)
            scanf("%d", &l[i]);
        for(int i = 1; i <= G; i++)
            scanf("%d", &w[i]);
        memset(dp, 0, sizeof(dp));  //達到每個狀態的方法數初始化為0
        dp[0][7500] = 1; // 0對應7500，掛上前0個鉤碼後，天枰達到平衡狀態7500的方法有1個，就是兩端都不掛
        for(int i = 1; i <= G; i++) {
            for(int j = 0; j <= 15000; j++) {
                if(dp[i-1][j]) { // 當放入i-1個鉤碼時狀態j已經出現且被統計過方法數，則直接使用統計結果，否則忽略當前狀態j
                    for(int k = 1; k <= C; k++) {
                        dp[i][j + w[i] * l[k]] += dp[i - 1][j];
                    }
                }
            }
        }
        printf("%d\n", dp[G][7500]);
    }
    return 0;
}