UVa839 Not so Mobile

 

我的解法： 建树，递归判断

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;

struct Node {
        Node() {
                wl=wr=dl=dr=0;
                l=r=0;
        }
        int wl;
        int dl;
        int wr;
        int dr;
        Node* l;
        Node* r;
};

Node* build()
{
        int wl, wr, dl, dr;
        scanf("%d%d%d%d", &wl, &dl, &wr, &dr);
        Node* root=new Node;
        if(wl==0)
        {
                Node* left=build();
                wl=left->wl + left->wr;
                root->l=left;
        }
        if(wr==0)
        {
                Node* right=build();
                wr=right->wl + right->wr;
                root->r=right;
        }

        root->wl=wl;
        root->wr=wr;
        root->dl=dl;
        root->dr=dr;
        return root;
}

bool equilibrium(Node* root)
{
        if(!root)
                return true;
        bool el=equilibrium(root->l);
        bool er=equilibrium(root->r);
        if(el&&er)
        {
                return (root->wl * root->dl == root->wr * root->dr);
        }
        else
        {
                return false;
        }
}



int main()
{
#ifndef ONLINE_JUDGE
        freopen("./uva839.in", "r", stdin);
#endif
        int T;
        scanf("%d", &T);
        while(T--) {
                Node* root=build();
                if(equilibrium(root))
                        printf("YES\n");
                else
                        printf("NO\n");

                if(T!=0)
                        printf("\n");
        }


    return 0;
}

解答解法：

// UVa839 Not so Mobile
// Rujia Liu
// 题意：输入一个树状天平，根据力矩相等原则判断是否平衡。采用递归方式输入，0表示中间结点
// 算法：在“建树”时直接读入并判断，并且无须把树保存下来
#include<iostream>
using namespace std;

// 输入一个子天平，返回子天平是否平衡，参数W修改为子天平的总重量
bool solve(int& W) {
  int W1, D1, W2, D2;
  bool b1 = true, b2 = true;
  cin >> W1 >> D1 >> W2 >> D2;
  if(!W1) b1 = solve(W1);
  if(!W2) b2 = solve(W2);
  W = W1 + W2;
  return b1 && b2 && (W1 * D1 == W2 * D2);
}

int main() {
  int T, W;
  cin >> T;
  while(T--) {
    if(solve(W)) cout << "YES\n"; else cout << "NO\n";
    if(T) cout << "\n";
  }
  return 0;
}