POJ 3628 Bookshelf 2(0-1背包)

叁歲伎倆 2024-02-17 19:31 3阅读 0赞

Bookshelf 2

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;9925</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;4462</td> 
  </tr> 
 </tbody> 
</table>

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 *Hi* (1 ≤ *Hi* ≤ 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: *Hi*

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 16
    3
    1
    3
    5
    6

Sample Output

AC代码：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    const int maxn=1e7+10;
    int dp[maxn];
    int value[maxn];
    int main(){
    	int N,B;
    	while(scanf("%d%d",&N,&B)==2){
    		int sum=0;
    		for(int i=1;i<=N;i++)
    		 {	
    		 scanf("%d",&value[i]);
    		 sum+=value[i];
    		 }
    		memset(dp,0,sizeof(dp));
    		for(int i=1;i<=N;i++){
    			for(int j=sum;j>=0;j--){
    				if(j>=value[i])
    				 dp[j]=max(dp[j],dp[j-value[i]]+value[i]);
    			}
    		}
    		
    		for(int i=0;i<=sum;i++){
    			if(dp[i]>=B){
    				printf("%d\n",abs(B-dp[i]));
    				break;
    			}
    		}
    	}
    	return 0;
    }

Bookshelf

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;8579</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;4306</td> 
  </tr> 
 </tbody> 
</table>

Description

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

Each of the *N* cows (1 ≤ *N* ≤ 20,000) has some height of *Hi* (1 ≤ *Hi* ≤ 10,000) and a total height summed across all *N* cows of *S*. The bookshelf has a height of *B* (1 ≤ *B* ≤ *S* < 2,000,000,007).

To reach the top of the bookshelf taller than the tallest cow, 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. Since more cows than necessary in the stack can be dangerous, your job is to find the set of cows that produces a stack of the smallest number of cows possible such that the stack can reach the bookshelf.

Input

\* Line 1: Two space-separated integers: *N* and *B*  
\* Lines 2..*N*\+1: Line *i*\+1 contains a single integer: *Hi*

Output

\* Line 1: A single integer representing the size of the smallest set of cows that can reach the bookshelf.

Sample Input

6 40
    6
    18
    11
    13
    19
    11

Sample Output

分析：排序+贪心

AC代码;

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=20000+10;
    int value[maxn]; 
    int main(){
    	int N,B;
    	while(scanf("%d%d",&N,&B)==2){
    		int sum=0;
    		for(int i=0;i<N;i++){
    			scanf("%d",&value[i]);
    			sum+=value[i];
    		}
    		sort(value,value+N);
    		int ans=N;
    		for(int i=0;i<N;i++){
    			if((sum-=value[i])>=B)ans--;
    			else break;
    		}
    		printf("%d\n",ans);
    	}
    	return 0;
    }