广度优先搜索
- 思路：
- 题目梳理
- 输入信息里包含 A / B = val；
- 输入一堆 A' / B' 需要求出它们的比值；
- 以操作数为节点，构建权重图，节点对应了关联节点及其比值；
- 可以进行简单的处理，将代数变量映射为整数数值，着上述图可以的节点键值转化为数组的下标；
-
for (int i = 0; i < size; ++i) {if (variables.find(equations[i][0]) == variables.end()) {variables[equations[i][0]] = nvars++;}if (variables.find(equations[i][1]) == variables.end()) {variables[equations[i][1]] = nvars++;}
}std::vector<std::vector<std::pair<int, double>>> digraph(nvars);
for (int i = 0; i < size; ++i) {int nodeA = variables[equations[i][0]];int nodeB = variables[equations[i][1]];digraph[nodeA].push_back(std::make_pair(nodeB, values[i]));digraph[nodeB].push_back(std::make_pair(nodeA, 1.0 /values[i]));
}
- 遍历求解的代数式：
- 如果变量不在已知的哈希表中，则结果为 -1.0；
- 如果两个变量相等，则结果为 1.0；
- 否则就可以从起点出发，通过广度优先搜索的方式，不断更新起点与当前点之间的路径长度，直到搜索到终点为止；
-
// bfs
std::queue<int> qu;
qu.push(na);
// result for search ratio
std::vector<double> ratio(nvars, -1.0);
ratio[na] = 1.0;
while (!qu.empty() && ratio[nb] < 0) {int x = qu.front();qu.pop();for (const auto [y, v] : digraph[x]) {if (ratio[y] < 0) {ratio[y] = ratio[x] * v;qu.push(y);}}
}val = ratio[nb];
- 完整代码
class Solution {
public:vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {int nvars = 0;std::unordered_map<std::string, int> variables;int size = equations.size();for (int i = 0; i < size; ++i) {if (variables.find(equations[i][0]) == variables.end()) {variables[equations[i][0]] = nvars++;}if (variables.find(equations[i][1]) == variables.end()) {variables[equations[i][1]] = nvars++;}}std::vector<std::vector<std::pair<int, double>>> digraph(nvars);for (int i = 0; i < size; ++i) {int nodeA = variables[equations[i][0]];int nodeB = variables[equations[i][1]];digraph[nodeA].push_back(std::make_pair(nodeB, values[i]));digraph[nodeB].push_back(std::make_pair(nodeA, 1.0 /values[i]));}std::vector<double> result;for (const auto & q : queries) {double val = -1.0;// all of the var existif (variables.find(q[0]) != variables.end() &&variables.find(q[1]) != variables.end()) {// node value from the graphint na = variables[q[0]];int nb = variables[q[1]];if (na == nb) {val = 1.0;} else {// bfsstd::queue<int> qu;qu.push(na);// std::vector<double> ratio(nvars, -1.0);ratio[na] = 1.0;while (!qu.empty() && ratio[nb] < 0) {int x = qu.front();qu.pop();for (const auto [y, v] : digraph[x]) {if (ratio[y] < 0) {ratio[y] = ratio[x] * v;qu.push(y);}}}val = ratio[nb];}}result.push_back(val);}return result;}
};
)
)
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