【三种解法】Not so Mobile UVA

立志用最少的代码做最高效的表达

Before being an ubiquous communications gadget, a mobile was just a structure made of strings and wires suspending colourfull things. This kind of mobile is usually found hanging over cradles of small babies.
The figure illustrates a simple mobile. It is just a wire, suspended by a string, with an object on each side. It can also be seen as a kind of lever with the fulcrum on the point where the string ties the wire. From the lever principle we know that to balance a simple mobile the product of the weight of the objects by their distance to the fulcrum must be equal. That is Wl × Dl = Wr × Dr where Dl
is the left distance,
Dr is the right distance, Wl is the left weight and Wr is the right weight.
In a more complex mobile the object may be replaced by a sub-mobile, as shown in the next figure.
In this case it is not so straightforward to check if the mobile is balanced so we need you to write a program that, given a description of a mobile as input, checks whether the mobile is in equilibrium or not.

Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The input is composed of several lines, each containing 4 integers separated by a single space.
The 4 integers represent the distances of each object to the fulcrum and their weights, in the format:
Wl Dl Wr Dr
If Wl or Wr is zero then there is a sub-mobile hanging from that end and the following lines define the the sub-mobile. In this case we compute the weight of the sub-mobile as the sum of weights of all its objects, disregarding the weight of the wires and strings. If both Wl and Wr are zero then the following lines define two sub-mobiles: first the left then the right one.

Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
Write ‘YES’ if the mobile is in equilibrium, write ‘NO’ otherwise.

Sample Input
1
0 2 0 4
0 3 0 1
1 1 1 1
2 4 4 2
1 6 3 2
Sample Output
YES

思路分析

经过观察很容易发现，本题的输入类似二叉树的先序遍历(即根左右)

以下三种解法为循序渐进的优化过程。

解法一：根据先序遍历建树，然后后序遍历判断天平是否平衡

解法二：在建树的同时判断天平是否平衡

解法三：隐式建树，通过递归的过程动态判断天平是否平衡。

解法一：先序建树、后序判断

#include<bits/stdc++.h>
using namespace std;
struct Node {int wl, wr, dl, dr;        //value Node *l = NULL, *r = NULL;    //left subtree、right subtree
};string s;
int T;
bool readInput(int& wl, int& dl, int &wr, int& dr) { //读取输入 if(getline(cin, s) && !s.empty()) {stringstream input(s);       //stringstream用法 input >> wl >> dl >> wr >> dr;return true;     //成功 }else return false;    //失败
}Node* createTree() {   //递归建树（先序） Node* root = NULL;int wl, dl, wr, dr;if(readInput(wl, dl, wr, d2)) {  //读入成功 root = new Node;root->wl = wl; root->wr = wr;root->dl = dl; root->dr = dr;if(wl == 0) root->l = createTree();if(wr == 0) root->r = createTree(); }return root;
}int dfs(Node* root) {  //后序遍历并判断是否平衡if(root == NULL) return -1;  //空树if(root->l != NULL) root->wl = dfs(root->l); //左子树if(root->r != NULL) root->wr = dfs(root->r);    //右子树if(root->wl == -1 || root->wr == -1) return -1;  //非法，剪枝if(root->wl * root->dl == root->wr*root->dr) return root->wl+root->wr;   //平衡，返回左右重量和else return -1;  //非法标志 }int main() {scanf("%d  ", &T);        //两个空格可吸收换行for(int i = 0; i < T; i++) {Node* bt = createTree();printf("%s%s\n", dfs(bt)!=-1 ? "YES" : "NO", i!=T-1?"\n":"");if(i != T-1) getchar(); //空行吸收 } return 0;
}

解法二：建树的同时判断

#include<bits/stdc++.h>
using namespace std;
struct Node{int Wl, Dl, Wr, Dr;     //数据域Node* left = nullptr, *right = nullptr;      //左右孩子指针
};int DFS(Node*&root, bool&f) {root = new Node();scanf("%d%d%d%d", &root->Wl, &root->Dl, &root->Wr, &root->Dr);if(root->Wl == 0)        //Wl是0root->Wl = DFS(root->left, f); //递归建立左子天平if(root->Wr == 0) root->Wr = DFS(root->right, f);if(root->Dl*root->Wl != root->Dr*root->Wr) //力矩不等f = false;return root->Wl + root->Wr;
}int main() {int n; scanf("%d", &n);for(int i = 0; i < n; i++) {Node *root = nullptr;bool f = true;DFS(root, f);printf("%s%s\n", i>0?"\n":"", f?"YES":"NO");}return 0;
}

解法三：隐式建树

#include<iostream>
using namespace std;bool solve(int &W) {        //此处W类似静态变量 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--) {cout << solve(W) ? "YES\n" : "NO\n";if(T) putchar('\n');}return 0;
}

择苦而安，择做而乐，虚拟终究不比真实精彩之万一

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【三种解法】Not so Mobile UVA - 839_19行代码AC

立志用最少的代码做最高效的表达

思路分析

解法一：先序建树、后序判断

解法二：建树的同时判断

解法三：隐式建树

择苦而安，择做而乐，虚拟终究不比真实精彩之万一

【三种解法】Not so Mobile UVA - 839_19行代码AC相关推荐

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