POJ 3628 Bookshelf 2（01背包）-白红宇

POJ 3628 Bookshelf 2（01背包）

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发布时间：2019-06-14

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Bookshelf 2

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9488		Accepted: 4311

Description

Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top.

FJ has N cows (1 ≤ N ≤ 20) each with some height of H_i (1 ≤ H_i ≤ 1,000,000 - these are very tall cows). The bookshelf has a height of B (1 ≤ B ≤ S, where S is the sum of the heights of all cows).

To reach the top of the bookshelf, one or more of the cows can stand on top of each other in a stack, so that their total height is the sum of each of their individual heights. This total height must be no less than the height of the bookshelf in order for the cows to reach the top.

Since a taller stack of cows than necessary can be dangerous, your job is to find the set of cows that produces a stack of the smallest height possible such that the stack can reach the bookshelf. Your program should print the minimal 'excess' height between the optimal stack of cows and the bookshelf.

Input

* Line 1: Two space-separated integers: N and B

* Lines 2..N+1: Line i+1 contains a single integer: H_i

Output

* Line 1: A single integer representing the (non-negative) difference between the total height of the optimal set of cows and the height of the shelf.

Sample Input

5 1631356

Sample Output

Source

一种做法是用dp[i]代表不超过i的可堆到最大程度，然后从m开始寻找第一个大于等于m的dp[i]就是答案，我用的是恰好装满的初始化条件，若可以出现一个恰好装满的解就输出……题目的S小于2000W其实比较大，1000W就差不多了

代码：

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             
              #include
              
               #include
               
                #include
                
                 #include
                 
                  #include
                  using namespace std;#define INF 0x3f3f3f3f#define CLR(x,y) memset(x,y,sizeof(x))#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)typedef pair
                   
                     pii;typedef long long LL;const double PI=acos(-1.0);const int N=10000010;int dp[N];int cow[30];int n,m;void zero_one_pack(int c,int w,int V){ for (int i=V; i>=c; --i) if(dp[i-c]+w>dp[i]) dp[i]=dp[i-c]+w;}int main(void){ int i,ans; while (~scanf("%d%d",&n,&m)) { CLR(dp,-INF); dp[0]=0; int sum=0; for (i=0; i
                    
                     =0) { printf("%d\n",i-m); break; } } } return 0;}

转载于:https://www.cnblogs.com/Blackops/p/5784459.html

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