POJ 1837 Balance (01背包)

2019-11-11 00:34:19

字體：大中小

供稿：網(wǎng)友

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

題意

有一個(gè)天平，左臂右臂各長(zhǎng)15，然后給出c，g。

c代表有幾個(gè)掛鉤，掛鉤給出負(fù)數(shù)代表在左臂的距離，正數(shù)則在右臂；

g代表有幾個(gè)砝碼，要你求出使得這個(gè)天平保持平衡有幾種方法，要求所有砝碼全部使用完。

思路

dp[i][j] $dp[i][j]$ ， i $i$ 為放置的砝碼數(shù)量， j $j$ 為平衡狀態(tài)，當(dāng) j=0 $j=0$ 時(shí)平衡。

但是這樣定義的話 j $j$ 有可能為負(fù)數(shù)，在當(dāng)作數(shù)組下標(biāo)的時(shí)候是不允許的，于是我們計(jì)算出天枰左傾或者右傾的最大限度是7500。

則 dp[i][7500] $dp[i][7500]$ 代表 i $i$ 個(gè)砝碼平衡的方法數(shù)目。

轉(zhuǎn)換為了一個(gè)01背包的問(wèn)題

狀態(tài)轉(zhuǎn)移方程： dp[i][j+w[i]?c[k]]+=dp[i?1][j] $dp[i][j+w[i]*c[k]]+=dp[i-1][j]$

對(duì)于第 i $i$ 個(gè)砝碼，它可以被在放置在任何地方，于是對(duì)每一個(gè)掛鉤處 j+w[i]?c[k] $j+w[i]*c[k]$ 都有貢獻(xiàn)。

最終答案： dp[g][7500] $dp[g][7500]$

AC 代碼

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>using namespace std;#include<vector>#include<queue>int dp[25][16000];int a[25],b[25];int main(){ int c,g; while(~scanf("%d%d",&c,&g)) { memset(dp,0,sizeof(dp)); for(int i=1; i<=c; i++) scanf("%d",a+i); for(int i=1; i<=g; i++) scanf("%d",b+i); dp[0][7500]=1; for(int i=1; i<=g; i++) for(int j=0; j<15000; j++) if(dp[i-1][j]) //一個(gè)小小的優(yōu)化 for(int k=1; k<=c; k++) dp[i][j+b[i]*a[k]]+=dp[i-1][j]; printf("%d/n",dp[g][7500]); } return 0;}

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