UVA 839 Not so Mobile

题目链接：http://acm.hust.edu.cn/vjudge/problem/19486

题意: 给你一个杠杆两端的物体的质量和力臂，如果质量为0，则下面是一个杠杆，
 当wl=wr=0,会先描述左子天平，然后是右子天平。大天平的W 是小天平两边的W总和
 
 天平的平衡条件 Wl * Dl == Wr * Dr，判断所有天平是否平衡。

先建根节点，然后判断左右结点质量是否为0，若是，则继续建树。在建完树回溯时判断当前天平是否平衡 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <cmath>
#include <stack>
#include <string>
#include <sstream>
#include <map>
#include <set>
#define pi acos(-1.0)
#define LL long long
#define ULL unsigned long long
#define inf 0x3f3f3f3f
#define INF 1e18
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define debug(a) printf("---%d---\n", a)
#define mem0(a) memset(a, 0, sizeof(a))
#define memi(a) memset(a, inf, sizeof(a))
#define mem1(a) memset(a, -1, sizeof(a))
using namespace std;
typedef pair<int, int> P;
const double eps = 1e-10;
const int maxn = 1e6 + 5;
const int mod = 1e8;

int failed, node;
struct Node{
int val;
}tree[maxn];
int NewNode(){
tree[++node].val = 0;
return node;
}
void solve(int rt)
{
int wl, dl, wr, dr;
int rt1 = NewNode(); // 左结点
int rt2 = NewNode(); // 右节点
cin >> wl >> dl >> wr >> dr;
if (!wl) solve(rt1); // 继续建树
else tree[rt1].val = wl;
if (!wr) solve(rt2);
else tree[rt2].val = wr;
if (tree[rt1].val * dl != tree[rt2].val * dr) // 天平不平衡
failed = 1;
tree[rt].val = tree[rt1].val + tree[rt2].val; // 质量和
}
int main(void)
{
// freopen("C:\\Users\\wave\\Desktop\\NULL.exe\\NULL\\in.txt","r", stdin);
int T, rt;
cin >> T;
while (T--){
failed = node = 0;
rt = NewNode();
solve(rt);
if (failed) puts("NO");
else puts("YES");
if (T) puts("");
}

return 0;
}