二进制枚举二叉树子集

#include <iostream>
#include <cstring>
#include <vector>
using namespace std;

const int MAX = 6;

struct Tree{
    double L, R;    //树的左边和右边的长度
    Tree():L(0), R(0){}
};

double width, w[MAX], sum[1<<MAX];
int vis[1<<MAX];
vector<Tree> tree[1<<MAX];        //枚举出来的结点可以组合成多个树

void dfs(int subset);        //枚举二叉树

int main(){
    freopen("input.txt", "r", stdin);
    int T;
    cin >> T;
    
    while(T--){
        int s;
        scanf("%lf%d", &width, &s);
        for(int i=0; i<s; i++){
            scanf("%lf", &w[i]);
        }
        
        //初始化 
        memset(sum, 0, sizeof(sum));
        for(int i=1; i<(1<<s); i++){
            tree[i].clear();    //清空树
            int k = 1;
            for(int j=0; j<s; j++){    //算出当前集合所有结点的重量和
                if(k & i){
                    sum[i] += w[j];
                }
                k = (k << 1);
            } 
        }
        
        //进行枚举
        int root = (1<<s) - 1;
        memset(vis, 0, sizeof(vis));
        dfs(root);
        
        //得到最大的答案
        double ans = -1;
        for(int i=0; i<tree[root].size(); i++){
            ans = max(tree[root][i].L + tree[root][i].R, ans);
        } 
        printf("%.10lf\n", ans);
        
    }
    
} 

void dfs(int subset){
    if(vis[subset])
        return;
    vis[subset] = 1;
    
    bool hasChild = false;
    
    for(int left = ((subset-1) & subset); left > 0; left = ((left-1) & subset)){    //枚举集合 
        hasChild = true;
        
        int right = subset ^ left;        //left 的补集
        
        double l = sum[right] / sum[subset];    //左臂长度
        double r = sum[left] / sum[subset];        //右臂长度
        
        dfs(left);    dfs(right);        //把左右树的子树也枚举出来
        for(int i=0; i<tree[left].size(); i++){            //将枚举出的左右子树组合起来 
            for(int j=0; j<tree[right].size(); j++){
                Tree t;
                t.L = max(tree[left][i].L + l, tree[right][j].L - r);
                t.R = max(tree[right][j].R + r, tree[left][i].R - l);
                if(t.L + t.R < width){                    //如果满足条件，枚举结果加一 
                    tree[subset].push_back(t); 
                }
            }
        }
    }
    
    if(!hasChild){        //叶子结点的左右臂为零 
        tree[subset].push_back(Tree());
    }
    
}

标签：subset,right,int,double,sum,枚举,子集,二叉树,left
来源： https://www.cnblogs.com/donkey9/p/15394650.html