Leetcode 399. Evaluate Division[medium]

题目：
Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.

Example:
Given a / b = 2.0, b / c = 3.0.
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .
return [6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is: vector < pair < string, string> > equations, vector& values, vector < pair < string, string> > queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector.

According to the example above:

equations = [ [“a”, “b”], [“b”, “c”] ],
values = [2.0, 3.0],
queries = [ [“a”, “c”], [“b”, “a”], [“a”, “e”], [“a”, “a”], [“x”, “x”] ].
The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

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这个题还是挺有意思的。
将每个分数当做图中的一个节点，边权为乘数的权值。
基础节点共有2n个（输入条件为n个分数），即这n个分数和这n个分数的倒数。
图中所有的新增节点的产生只有两种途径：
1.基础节点（基础分数）*基础分数（边）得到一个新增节点（新增分数）
2.新增节点（新增分数）*基础分数（边）得到一个新增节点（新增分数）

也就是说，新增节点是不可以作为边的，因为即便你想让新增节点作为边，这个分数的值在dfs之前是不知道的，所以也没办法作为边。

class Solution {    #define N 11111    #define M 111111public:    map< pair<string, string>, int> mp;    double ans[N];    bool vis[N];    int cnt;    int ts;    int head[N], nxt[M], to[M];    double len[M];    void add(int u, int v, double w) {        to[ts] = v; len[ts] = w; nxt[ts] = head[u]; head[u] = ts++;        //cout << u << "   " << v << "   " << w << endl;    }    vector<double> calcEquation(vector< pair<string, string> > equations, vector<double>& values, vector< pair<string, string> > queries) {        memset(head, -1, sizeof head);        memset(vis, 0, sizeof vis);        mp.clear();        cnt = 0;        ts = 0;        pair<string, string> pr;        int n = equations.size();        for (int i = 0; i < n; i++)             if (mp[equations[i]] == 0) {                mp[equations[i]] = ++cnt;                add(0, mp[equations[i]], values[i]);//multiply                pr.first = equations[i].second;                pr.second = equations[i].first;                equations.push_back(pr);                values.push_back(1.0 / values[i]);                if (mp[pr] == 0) mp[pr] = ++cnt;                add(0, mp[pr], 1.0 / values[i]);//divide            }        int m = equations.size();        for (int i = 0; i < equations.size(); i++) {//注意这两层循环             for (int j = 0; j < m; j++) {                if (i == j) continue;                if (equations[i].first == equations[j].second) {//multiply                    pr.first = equations[j].first;                    pr.second = equations[i].second;                    if (mp[pr] == 0) {                        mp[pr] = ++cnt;                        equations.push_back(pr);//新增                     }                    add(mp[equations[i]], mp[pr], values[j]);                }                if (equations[i].second == equations[j].first) {//multiply                    pr.first = equations[i].first;                    pr.second = equations[j].second;                    if (mp[pr] == 0) {                        mp[pr] = ++cnt;                        equations.push_back(pr);//新增                     }                    add(mp[equations[i]], mp[pr], values[j]);                }            }        }         ans[0] = 1.0;        vis[0] = true;        dfs(0);        vector<double> rt;        for (int i = 0; i < queries.size(); i++) {            if (mp[queries[i]] == 0) rt.push_back(-1.0);            else rt.push_back(ans[mp[queries[i]]]);        }        return rt;    }    void dfs(int x) {        for (int i = head[x]; ~i; i = nxt[i]) {            if (!vis[to[i]]) {                ans[to[i]] = ans[x] * len[i];                vis[to[i]] = true;                dfs(to[i]);            }         }    }};

这题在写测试代码的时候挺恶心的，把测试代码也放上来吧。

int main() {    Solution s;    vector< pair<string, string> > vequations;    vequations.push_back(make_pair("x1", "x2"));    vequations.push_back(make_pair("x2", "x3"));    vequations.push_back(make_pair("x3", "x4"));    vequations.push_back(make_pair("x4", "x5"));    vector<double> values;    values.push_back(3.0);    values.push_back(4.0);    values.push_back(5.0);    values.push_back(6.0);    vector< pair<string, string> > queries;    queries.push_back(make_pair("x1", "x5"));    queries.push_back(make_pair("x5", "x2"));    queries.push_back(make_pair("x2", "x4"));    queries.push_back(make_pair("x2", "x2"));    queries.push_back(make_pair("x2", "x9"));    vector<double> ans;    ans = s.calcEquation(vequations, values, queries);    for (int i = 0; i < ans.size(); i++) cout << ans[i] << endl;    return 0;}