C. Anji's Binary Tree

网站：https://codeforces.com/problemset/problem/1900/C
这里比较容易搞混的就是各个节点的关系，不要用2*n来表示左孩子！！
以及记得添加快读快写

ios::sync_with_stdio(false);
cin.tie(nullptr);

加在int main（）里即可
这里还有一个priority_queue的小技巧：结构体内部定义运算符，好像和sort的cmp有不同？

#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
typedef long long ll;
using namespace std;


struct leet
{
	int left, right;
	char tar;
	leet() { left = 0; right = 0; tar = ' '; }
};
leet node[300100];
struct dat
{
	int val, times;
	dat(){}
	dat(int va, int tim) { val = va; times = tim; }
	bool operator<(const dat& dl) const
	{
		return this->times > dl.times;//※指把小的times放前面
	}
};

int main()
{
	ios::sync_with_stdio(false);//imp
	cin.tie(nullptr);//imp
	int t; cin >> t;
	for (int ii = 0; ii < t; ii++)
	{
		//queue<dat>qd;
		priority_queue<dat, vector<dat>>qd;//用优先队列让times少的先过，类似dij？(
		int n; cin >> n; cin.get();
		string s; getline(cin, s);
		for (ll i = 1; i <= n; i++)
		{
			int xx, yy; cin >> xx >> yy;
			node[i].tar = s[i - 1];
			node[i].left = xx;
			node[i].right = yy;
		}
		qd.push(dat(1, 0));//排入第一个节点
		ll mintimes = 0, jd = 1;
		bool isend = 0;
		while (!isend)//无所谓的，写一个whiletrue也是一样
		{
			dat temp = qd.top();
			qd.pop();
			mintimes = temp.times;
			jd = temp.val;
			if (node[jd].left == 0 and node[jd].right == 0)
			{
				isend = 1;
				break;
			}
			if (node[jd].tar == 'U')//如果是u那么times一定得加一，然后分别遍历左右
			{
				if (node[jd].left != 0)qd.push(dat(node[jd].left, mintimes + 1));
				if (node[jd].right != 0)qd.push(dat(node[jd].right, mintimes + 1));
			}
			else if (node[jd].tar == 'L')//本来就是左边那么遍历左节点就不用加一，遍历右节点才需要加一
			{
				if (node[jd].left != 0)qd.push(dat(node[jd].left, mintimes));
				if (node[jd].right != 0)qd.push(dat(node[jd].right, mintimes + 1));
			}
			else//同上
			{
				if (node[jd].right != 0)qd.push(dat(node[jd].right, mintimes));
				if (node[jd].left != 0)qd.push(dat(node[jd].left, mintimes + 1));
				
			}
		}
		cout << mintimes << endl;

	}
	return 0;
}

♥对了，爱你rzbb♥