洛谷P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper

P3052 [USACO12MAR]摩天大楼里的奶牛Cows in a Skyscraper

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)

输入输出格式

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：

• Line 1: A single integer, R, indicating the minimum number of elevator rides needed.

• Lines 2..1+R: Each line describes the set of cows taking

one of the R trips down the elevator. Each line starts with an integer giving the number of cows in the set, followed by the indices of the individual cows in the set.

输入输出样例

输入样例#1：

4 10
5
6
3
7

输出样例#1：

3
2 1 3
1 2
1 4

说明

There are four cows weighing 5, 6, 3, and 7 pounds. The elevator has a maximum weight capacity of 10 pounds.

We can put the cow weighing 3 on the same elevator as any other cow but the other three cows are too heavy to be combined. For the solution above, elevator ride 1 involves cow #1 and #3, elevator ride 2 involves cow #2, and elevator ride 3 involves cow #4. Several other solutions are possible for this input.

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 20
int n,m,a[maxn],b[maxn],mid;
bool flag=;
void dfs(int pos,int num){
    if(pos==n+){
        flag=;
        return;
    }
    if(flag)return;
    for(int i=;i<=num;i++){
        if(m-b[i]>=a[pos]){
            b[i]+=a[pos];
            dfs(pos+,num);
            b[i]-=a[pos];
        }
    }
    if(num==mid)return;
    b[num+]=a[pos];
    dfs(pos+,num+);
    b[num+]=;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=;i<=n;i++)scanf("%d",&a[i]);
    int ans=n,l=,r=n;
    while(l<=r){
        mid=(l+r)>>;
        memset(b,,sizeof(b));
        flag=;
        dfs(,);
        if(flag)ans=mid,r=mid-;
        else l=mid+;
    }
    printf("%d",ans);
}

100分 迭代加深搜索（二分深度）

/*
    f 数组为结构体
    f[S].cnt 表示集合 S 最少的分组数
    f[S].v 表示集合 S 最少分组数下当前组所用的最少容量
    f[S] = min(f[S], f[S - i] + a[i]) (i ∈ S)
    运算重载一下即可。
*/
#include<cstdio>
#include<iostream>
int n,m,w;
int a[];
struct node{
    int cnt,v;
    node operator + (const int b)const{
        node res;
        if(v+b<=w)res.cnt=cnt,res.v=v+b;
        else res.cnt=cnt+,res.v=b;
        return res;
    }
    bool operator < (const node b)const{
        if(cnt==b.cnt)return v<b.v;
        return cnt<b.cnt;
    }
}f[<<];
node min(node x,node y){
    if(x<y)return x;
    return y;
}
node make_node(int x,int y){
    node res;
    res.cnt=x;res.v=y;
    return res;
}
int main(){
    int sta;
    scanf("%d%d",&n,&w);
    m=(<<n)-;
    for(int i=;i<=n;i++)scanf("%d",&a[i]);
    for(sta=;sta<=m;sta++){
        f[sta]=make_node(1e9,w);
        for(int i=;i<=n;i++){
            if(!((<<i-)&sta))continue;
            f[sta]=min(f[sta],f[(<<i-)^sta]+a[i]);
        }
    }
    printf("%d",f[m].cnt+);
    return ;
}

100分 状压dp