59 lines
1.3 KiB
C++
59 lines
1.3 KiB
C++
#include <iostream>
|
||
//前序遍历 递归的天花板
|
||
using namespace std;
|
||
struct mobile{
|
||
double lw, ll, rw, rl;
|
||
double weight;
|
||
mobile* left;
|
||
mobile* right;
|
||
|
||
mobile(double a, double b, double c, double d) : lw(a), ll(b), rw(c), rl(d), left(nullptr), right(nullptr) {}
|
||
};
|
||
|
||
bool solve(mobile* &x){
|
||
double a, b, c, d;
|
||
bool b1 = true, b2 = true;
|
||
cin >> a >> b >> c >> d;
|
||
x = new mobile(a, b, c, d);
|
||
if(!x->lw) b1 = solve(x->left);
|
||
if(!x->rw) b2 = solve(x->right);
|
||
double leftWeight = x->left ? x->left->weight : x->lw;
|
||
double rightWeight = x->right ? x->right->weight : x->rw;
|
||
x->weight = leftWeight + rightWeight;
|
||
//关注return如果再solve一遍那就是多次递归了，不能再solve了
|
||
return b1 && b2 && (leftWeight * x->ll == rightWeight * x->rl);
|
||
|
||
}
|
||
|
||
|
||
|
||
int main() {
|
||
ios::sync_with_stdio(false);
|
||
cin.tie(nullptr);
|
||
|
||
int n;
|
||
cin >> n;
|
||
string tmp;
|
||
getline(cin, tmp); // 读取第一行后的换行
|
||
|
||
for (int i = 0; i < n; i++) {
|
||
mobile* root = nullptr;
|
||
bool ok = solve(root);
|
||
cout << (ok ? "YES" : "NO") << "\n";
|
||
|
||
if (i != n - 1) cout << "\n"; // 两个查询之间空一行
|
||
}
|
||
/*
|
||
for(int i = 0; i < n - 1; i++){
|
||
cin >> a >> b >> c >> d;
|
||
mobile* tmp = new mobile(a, b, c, d);
|
||
if(a && !b){
|
||
|
||
}
|
||
}
|
||
这是不好的，应该是自然而然地形成一棵树
|
||
*/
|
||
|
||
return 0;
|
||
}
|