題干：

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

題目大意：

? ? 其實很簡單啦就是給你n個高度，讓你挑其中的幾個摞起來。要求摞起來以后的總高度超過b，問你這樣的高度的最小值是多少。（讓你輸出 ?最小值 - b ）

有n頭牛，已知每頭牛的高度和書架高度，一頭牛可以站在另一頭牛身上，總高度是他們的高度之和。要求能夠達到書架的頂端，即這些牛的總高度不低于書架高度，求滿足條件的總高度的最小值，輸出他們的差值。

解題報告：

? ? 這題高度小于100W(1e6)，但是題目說b要小于總高度和，n<=20，也就是說b<=2000W(2e7)，這個數字其實做0-1背包就已經很牽強了（我也試過了開了2e7的數組然后MLE了），不過還好這題數據量沒給到這么大，開個100W的數組就過去了。不過要說到正解，這題是搜索啊！n<=20這么小的數據量你不去dfs嗎？！

下面附四個代碼：

AC代碼1：（自己寫 ?裝滿類型0-1背包）

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm>using namespace std; const int MAX = 1000000 +5; const int INF = 0x3f3f3f3f; int dp[MAX]; int v[50];bool cmp(const int a,const int b) {return a>b; } int main() {int n,b;cin>>n>>b;for(int i = 1; i<=n; i++) {cin>>v[i];}sort(v+1,v+n+1,cmp);//cout <<"kaishile"<<endl;memset(dp,INF,sizeof(dp));dp[0] = 0;for(int i = 1; i<=n; i++) {for(int j = b + v[1]; j>=v[i]; j-- ) {dp[j] = min(dp[j],dp[j -v[i] ] + 1);}}int ans = -1;for(int i = b; i<=b+v[1]; i++) {if(dp[i] !=INF) {ans = i - b; break;}}cout << ans<<endl;return 0 ; }

AC代碼2：（網絡版的 裝滿類型0-1背包）?

#include<cstdio> using namespace std; const int mm=20000000; bool f[mm]; int h[22]; int i,j,k,n,b,m; int main() {while(scanf("%d%d",&n,&b)!=-1){for(m=i=0;i<n;++i)scanf("%d",&h[i]),m+=h[i];for(i=0;i<=m;++i)f[i]=0;f[0]=1;for(i=0;i<n;++i)for(j=m-h[i];j>=0;--j)if(f[j])f[j+h[i]]=1;for(i=b;i<=m;++i)if(f[i])break;printf("%d\n",i-b);}return 0; }

上面這個代碼看起來0-1背包不太一樣，其實是一樣的。只不過一般j從m開始遞減，這里是從m-h[i]開始的，其實是一樣的。

之所以要把這個代碼貼上來其實主要不是這個地方，而是他只用了bool類型的dp數組，因為這里只需要記錄這個高度是否可以到達，而不需要知道具體能到達的高度是多少。其實往深里講，這是因為這個題只有h[i] 這一個權值，而不是標準的0-1背包還需要w和v兩個權值，所以其實就算是按照我上面自己寫的代碼那樣寫，其實每個dp數組中只要不是-INF，那么這個值一定就是高度一定就是i（也就是dp[i]一定等于i），所以其實這兩種寫法是等價的。

AC代碼3：（純裸的0-1背包）

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int dp[2000002], h[22]; int main() {int n, m, i, j;while(~scanf("%d%d",&n,&m)){int sum = 0;memset(dp,0,sizeof(dp));for(i = 1; i <= n; i++){scanf("%d",&h[i]);sum += h[i];}for(i = 1; i <= n; i++)for(j = sum; j >= h[i]; j--)dp[j] = max(dp[j], dp[j - h[i]] + h[i]);int Min = sum;for(i = m; i <= sum; i++)if(dp[i] >= m && dp[i] - m < Min)Min = dp[i] - m;printf("%d\n",Min);}return 0; }

AC代碼4：（網絡版 dfs）

#include<cstdio> #include<cstring> #include<iostream> using namespace std; int h[22], ans, flag; int n, m; void dfs(int k, int s) {if(s == m){ans = 0;return ;}if(s >= m){if(s - m < ans)ans = s - m;return ;}for(int i = k; i < n; i++){dfs(i+1,s+h[i]);} } int main() {int i;while(cin >> n >> m){int sum = 0;flag = 0;for(i = 0; i < n; i++){cin >> h[i];sum += h[i];}if(sum == m){cout << "0" << endl;continue;}ans = sum;dfs(0,0);cout << ans << endl;}return 0; }

?

總結

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