枚舉二叉樹然後計算其寬度即可，每次枚舉兩個節點構造一個父節點，計算寬度時需要注意的是每棵樹的左節點的右邊緣可能超過其右子樹的左邊緣，反之亦然。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 12;
int n,s,idx,w[maxn],vis[maxn];
double r,res,lw[maxn],rw[maxn];
void dfs(int cnt){
if(cnt == s) return;
for(int i = 0; i < maxn; i++){
if(!vis[i]) continue;
for(int j = 0; j < maxn; j++){
if(!vis[j] || i == j) continue;
double L = max(lw[i],lw[j]-1), R = max(rw[j],rw[i]-1);
if(1 + L + R < r){
vis[i] = vis[j] = 0;
if(cnt == s - 1) res = max(res,1+L+R);
int id = idx++;
vis[id] = 1;
w[id] = w[i] + w[j];
lw[id] = L + 1.0*w[j]/w[id];
rw[id] = R + 1.0*w[i]/w[id];
dfs(cnt+1);
vis[--idx] = 0;
vis[i] = vis[j] = 1;
}
}
}
}
int main()
{
// freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
scanf("%d",&n);
while(n--){
res = -1; idx = 0;
memset(vis,0,sizeof(vis));
memset(lw,0,sizeof(lw));
memset(rw,0,sizeof(rw));
scanf("%lf%d",&r,&s);
for(int i = 0; i < s; i++){
scanf("%d",&w[i]);
vis[idx++] = 1;
}
if(s == 1){ printf("%.10lf\n",0.0); continue; }
dfs(1);
printf("%.10lf\n",res);
}
return 0;
}