https://www.acwing.com/problem/content/1630/
思路:
思路不难,按照题目意思判断即可,但是这个dfs有些难写,值得学习,特记录此题。
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
const int N = 40;
int pre[N], in[N];
unordered_map<int, int> pos;
bool ans;
int build(int il, int ir, int pl, int pr, int& sum)
{
int root = pre[pl];
int k = pos[abs(root)];
if (k < il || k > ir)
{
ans = false;
return 0;
}
int left = 0, right = 0, ls = 0, rs = 0;
if (il < k) left = build(il, k - 1, pl + 1, pl + 1 + k - 1 - il, ls);
if (k < ir) right = build(k + 1, ir, pl + 1 + k - 1 - il + 1, pr, rs);
if (ls != rs) ans = false;
sum = ls;
if (root < 0)
{
if (left < 0 || right < 0) ans = false;
}
else sum ++ ;
return root;
}
int main()
{
int T;
cin >> T;
while (T -- )
{
int n;
cin >> n;
for (int i = 0; i < n; i ++ )
{
cin >> pre[i];
in[i] = abs(pre[i]);
}
sort(in, in + n);
pos.clear();
for (int i = 0; i < n; i ++ ) pos[in[i]] = i;
ans = true;
int sum;
int root = build(0, n - 1, 0, n - 1, sum);
if (root < 0) ans = false;
if (ans) puts("Yes");
else puts("No");
}
return 0;
}
标签:判断,int,sum,红黑树,il,ans,root,pl
From: https://www.cnblogs.com/xjtfate/p/16603241.html