状压dp最小分组

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#算法

本文介绍了一个关于奶牛下楼的趣味问题，奶牛们在攀登完摩天大楼后不愿走楼梯下楼，于是必须使用电梯。问题转化为如何在不超过电梯最大载重的情况下，用最少的电梯次数将所有奶牛送回地面。文章通过动态规划算法解决这个问题，给出了输入格式、输出格式，并展示了一个C++实现的解决方案。

题目描述

A little known fact about Bessie and friends is that they love stair climbing races. A better known fact is that cows really don't like going down stairs. So after the cows finish racing to the top of their favorite skyscraper, they had a problem. Refusing to climb back down using the stairs, the cows are forced to use the elevator in order to get back to the ground floor.

The elevator has a maximum weight capacity of W (1 <= W <= 100,000,000) pounds and cow i weighs C_i (1 <= C_i <= W) pounds. Please help Bessie figure out how to get all the N (1 <= N <= 18) of the cows to the ground floor using the least number of elevator rides. The sum of the weights of the cows on each elevator ride must be no larger than W.

给出n个物品，体积为w[i]，现把其分成若干组，要求每组总体积<=W，问最小分组。(n<=18)

输入格式

* Line 1: N and W separated by a space.

* Lines 2..1+N: Line i+1 contains the integer C_i, giving the weight of one of the cows.

输出格式

* A single integer, R, indicating the minimum number of elevator rides needed.

one of the R trips down the elevator.

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=20;
int dp[1<<maxn];
int g[1<<maxn];
int val[maxn];
int main()
{
    int n,w;cin>>n>>w;
    for(int i=0;i<n;i++) cin>>val[i];
    memset(dp,0x3f,sizeof(dp));
    dp[0]=1;
    g[0]=w;
    for(int i=0;i<(1<<n);i++)
    {
        for(int j=0;j<n;j++)
        {
            if((i>>j)&1) continue;
            if(g[i]>=val[j]&&dp[i|(1<<j)]>=dp[i])
            {
                dp[i|(1<<j)]=dp[i];
                g[i|(1<<j)]=max(g[i|(1<<j)],g[i]-val[j]);
            }
            else if(g[i]<val[j]&&dp[i|(1<<j)]>=dp[i]+1)
            {
                dp[i|(1<<j)]=dp[i]+1;
                g[i|(1<<j)]=max(g[i|(1<<j)],w-val[j]);
            }
        }
    }
    cout<<dp[(1<<n)-1]<<endl;
}