Mobile Computing UVA - 1354

模拟递归构建二叉树，然后计算每次递归构建平衡之后的长度，并且在所有节点都用完的时候与已知最大的长度比较大小，存储最大的长度。这个题目注意在计算新建平衡树之后的左端长和右端长度的的时候注意i和j的下标位置，自己推导吧：动力*动力臂=阻力*阻力臂，在这个上面卡了好久。其他的见如下代码：

#include<iostream>
#include<vector>
#include<string>
#include<set>
#include<stack>
#include<queue>
#include<map>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<deque>
using namespace std;

#define MAX 12

int n, s;
double r,lLength[MAX],rLength[MAX];
int nWeight[MAX];
bool visit[MAX];
double ans;
int ind;

void build_tree(int amount){
if (amount == s) return;
for (int i = 0; i < MAX; i++){
if (!visit[i]){
for (int j = 0; j < MAX; j++){
if (i != j&&!visit[j]){
double L = max(lLength[i],lLength[j]-1);
double R = max(rLength[i]-1,rLength[j]);
if (L + R + 1 < r){
if (amount == s - 1) ans = max(ans, L + R + 1);
visit[i] = true;
visit[j] = true;
visit[ind] = false;
nWeight[ind] = nWeight[i] + nWeight[j];
lLength[ind] = nWeight[j] * 1.0 / nWeight[ind] + L;
rLength[ind] = nWeight[i] * 1.0 / nWeight[ind] + R;
cout << nWeight[ind] << " " << lLength[ind] << " " << rLength[ind] << endl;
ind++;
build_tree(amount + 1);
ind--;
visit[ind] = true;
visit[i] = false;
visit[j] = false;
}
}
}
}
}
}

int main(){
cin >> n;
while (n--){
cin >> r >> s;
ind = s;
for (int i = 0; i < MAX; i++){
visit[i] = true;
lLength[i] = 0;
rLength[i] = 0;
}
for (int i = 0; i < s; i++){
cin >> nWeight[i];
visit[i] = false;
}
ans = -1;
if (s == 1) { printf("0.0000000000\n"); continue; }
build_tree(1);
printf("%.13lf\n", ans);

}
return 0;
}