[01背包]POJ 3628 Bookshelf 2
2014-02-01 09:13
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Bookshelf 2
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
Sample Output
Source
USACO 2007 December Bronze
解题报告:
此题为01背包。代码如下:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[1000005];
int a[10005];
int main(){
int n,m,sum;
while(scanf("%d%d",&n,&m)==2){
memset(dp,0,sizeof(dp));
sum=0;
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
sum+=a[i];
}
for(int i=0;i<n;i++)
for(int j=sum;j>=a[i];j--)
dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
for(int i=0;i<=sum;i++)
if(dp[i]>=m){
printf("%d\n",dp[i]-m);
break;
}
}
return 0;
}
Bookshelf 2
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6934 | Accepted: 3198 |
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
1
Source
USACO 2007 December Bronze
解题报告:
此题为01背包。代码如下:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[1000005];
int a[10005];
int main(){
int n,m,sum;
while(scanf("%d%d",&n,&m)==2){
memset(dp,0,sizeof(dp));
sum=0;
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
sum+=a[i];
}
for(int i=0;i<n;i++)
for(int j=sum;j>=a[i];j--)
dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
for(int i=0;i<=sum;i++)
if(dp[i]>=m){
printf("%d\n",dp[i]-m);
break;
}
}
return 0;
}
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