Mobile Computing UVA - 1354 枚举二叉树

问题

https://vjudge.net/problem/UVA-1354

分析

参考：https://www.cnblogs.com/worldcreator-zh/p/10769859.html
能力太差了，知道是枚举二叉树，很多细节还是处理不好。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <map>
#include <string>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;
const int maxn=6;
struct Tree{
    double L,R;
    Tree(double L=0,double R=0):L(L),R(R){}
};
int s,kase=0,vis[1<<maxn],w[maxn];
vector<Tree> tree[1<<maxn];  //tree[i]存储i对应的Tree结果
double r,ans,sum[1<<maxn];

//遍历subset可以构建的所有的子树，存入tree[subset中
void dfs(int subset){
    if(vis[subset]) return;
    vis[subset]=1;
    bool have_children=false;
    for(int L=(subset-1)&subset;L;L=(L-1)&subset){
        int R=subset^L;
        have_children=true;
        double d1=sum[R]/sum[subset],d2=sum[L]/sum[subset];
        dfs(L);
        dfs(R);
        for(int i=0;i<tree[L].size();++i){
            for(int j=0;j<tree[R].size();++j){
                Tree t;
                t.L=max(tree[L][i].L+d1,tree[R][j].L-d2);
                t.R=max(tree[L][i].R-d1,tree[R][j].R+d2);
                if(t.L+t.R<r) tree[subset].push_back(t);
            }
        }
    }
    //这个是必须的
    if(!have_children) tree[subset].push_back(Tree());
}

int main(void){
    scanf("%d",&kase);
    while(kase--){
        scanf("%lf%d",&r,&s);
        for(int i=0;i<s;++i){
            scanf("%d",&w[i]);
        }
        int S=1<<s;
        memset(vis,0,sizeof(vis));
        for(int i=0;i<S;++i){
            sum[i]=0;
            tree[i].clear();
            for(int j=0;j<s;++j){
                if(i&(1<<j)){
                    sum[i]+=w[j];
                }
            }
        }
        dfs(S-1);
        ans=-1;
        for(int i=0;i<tree[S-1].size();++i){
            ans=max(ans,tree[S-1][i].L+tree[S-1][i].R);
        }
        printf("%.10lf\n", ans);
    }
    return 0;
}