Algorithm: POJ 3628 Bookshelf 2 (DFS, 01 backpack)

The main effect of the topic:

There are n numbers (1~100w), and now there is a number b,1<=b<=s (S is the sum of n digits). To select some number combinations from n digits, the sum of the numbers and x of the combined number is greater than or equal to B, what is the smallest x-b of all x?

Ideas:

Just beginning to see this problem, found that the number is so large that the memory is not enough to use a backpack. and n Max only 20, so directly with dfs+ minus 0MS ...

Then with the backpack, opened the 2000w+ array, memset initialization, decisively mle ... Then looked at the next discuss, found that using memset will add a lot of extra memory.

So instead of initializing for the For loop, the memory is sufficient. And then there's the Naked 01 backpack.

Something:

The data is enough water, DP Open 20W array is over, the complexity of violent 2^20 is also no pressure

DFS + minus 0MS:

#include <iostream>  
#include <queue>  
#include <cstdio>  
#include <cstring>  
#include <cmath>  
      
typedef long long Int64;  
      
const int MAXN = 1000010;  
const int INF = 0X3F3F3F3F;  
int n, b;  
int h[25];  
int sum[25];  
int ans;  
      
void Dfs (int cur, int tol) {  
    if (cur>n+1 | | | ans==b) return;  
    if (tol >= ans) return;  
    if (tol >= b) {  
        ans = min (tol, ans);  
        return;  
    }  
    if (tol + sum[n]-sum[cur-1] < b) return  
        ;  
      
    DFS (cur+1, tol);  
    DFS (cur+1, tol+h[cur]);  
int main () {  
      
      
    while (~scanf ("%d%d", &n, &b)) {  
      
        sum[0] = 0;  
        for (int i=1; i<=n; ++i) {  
            scanf ("%d", &h[i]);  
            if (i==0) sum[0]=h[i];  
            else sum[i] = Sum[i-1]+h[i];  
        }  
      
        ans = INF;  
        DFS (1, 0);  
        printf ("%d\n", ans-b);  
    }  
      
    return 0;  
}