"Dynamic Programming · backpack" Gift

Title Description

answer

If the current first $i$ th item is not selected and not selected article is minimal, then this means that a volume of less than $a_i$When there will be no other items to fill a backpack, then finished conducted for other items 01 backpack after contributed answer is: $\sum_{i=m-a_i+1}^{m}f[i]$.

But doing so there are two difficulties: time complexity are not good enough, and can not guarantee whether the items are currently no election is the smallest of items.

So we enumerate whether a point is a minimum of time, make it a point not to force the election, it is smaller than both the election, it looks larger than pick cases.

We use to think with $O (n ^ 2)$level algorithm. Obviously, the answer is divided into two parts:

• Than this point to do 01 backpack big volume.
• Small size than this point forcibly selected.

At this point we need to sort descending, not to omit the first dimension state, we can get a large volume of the backpack portion 01 results.

Very simple transfer equation is: $f[i][j]=f[i-1][j]+f[i-1][j-a[i]]$

Then forcibly select one item, forced state transition according to the backpack 01 is formed larger items, other $S=\sum_{j=i+1}^{m} a_j$.

It can be transferred, set up $g[i]$ indicates a mandatory selection of $i+1$ after the program items: $g[j]=f[i-1][j-S]$

Then the answer is: $ans\ =\ \sum_{i=1}^{n} \sum_{j=m-a+i+1}^{m} g_j$

code show as below:

#include <bits/stdc++.h>

#define int long long

using namespace std;
const int N = 2000;
const int P = 1000000007;

int n, m, ans = 0;
int a[N], f[N][N], s[N];

signed main(void)
{
	cin>>n>>m;
	for (int i=1;i<=n;++i) cin>>a[i];
	sort(a+1,a+n+1);
	reverse(a+1,a+n+1);//从大到小排序 
	f[0][0] = 1;
	for (int i=1;i<=n;++i)
	    for (int j=0;j<=m;++j)
	    {
	        if (j >= a[i]) f[i][j] = (f[i-1][j]+f[i-1][j-a[i]]) % P;//01背包 
	        else f[i][j] = f[i-1][j];
	    }
	for (int i=n;i>=1;--i) s[i] = s[i+1]+a[i];//后缀和 
	for (int i=1;i<=n;++i)
	{
		//枚举强制不选择的物品
		//强制选择i+1-n的物品
		int v = s[i+1];
		int g[10000] = {};
		for (int j=m;j>=v;--j) g[j] = f[i-1][j-v];
		for (int j=m-a[i]+1;j<=m;++j) ans = (ans+g[j]) % P;
	}
	cout<<ans<<endl;
	return 0;
}

to sum up:

This title challenge is to think of how to answer a total and pass a minimum-related, then you can fool around various optimization.