[leetcode] 399. Evaluate Division

我是链接

看到这道题，2个点和一个权值，然后想到图，但是leetcode就是这样，没给数据范围，感觉写起来很费劲，然后就开始用图来做，添加边的时候，注意正向边和反向变，然后查询的时候，先判断2个点是否都出现，然后判断是不是自己到自己，最后用dfs或者bfs到图里面去搜索，第一次交的时候，忘记用vis标记访问过的点，导致tle，加上之后，就ac了，套路一般吧，不知道还有什么trick。

double dfs(vector<vector<pair<int, double> > > &e, int x, int y) {
    int n = e.size();
    vector<double> dis(n, 1);
    queue<int> q;
    q.push(x);
    vector<bool> tag(n, 0);
    while(!q.empty()) {
        int u = q.front(); q.pop();
        //cout << u << endl;
        for (int i = 0; i < e[u].size(); i++) {
            int a = e[u][i].first;
            double b = e[u][i].second;
            if(tag[a]) continue;
            dis[a] = dis[u] * b;
            if(a == y) return dis[a];
            q.push(a);
            tag[a] = 1;
        }
    }
    return -1.0;
}
   vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> query) {
        vector<vector<pair<int, double> > > e;

        map<string, int> m;
        for (int i = 0; i < equations.size(); i++) {
            string x = equations[i].first, y = equations[i].second;
            if(!m.count(x)) {
                vector<pair<int, double> > t;
                m[x] = e.size(); e.push_back(t);
            }
            if(!m.count(y)) {
                vector<pair<int, double> > t;
                m[y] = e.size(); e.push_back(t);
            }
            int a = m[x],b = m[y];
            e[a].push_back({b, values[i]});
            e[b].push_back({a, 1.0 / values[i]});
        }
        //cout << "ad" << endl;
        //cout << e.size() << endl;
        vector<double> res;
        for (int i = 0; i < query.size(); i++) {
            string x = query[i].first, y = query[i].second;
            //cout <<x << " " << y << endl;
            if(!m.count(x) || !m.count(y)) {
                res.push_back(-1.0);
            } else {
                if(x == y) {
                    res.push_back(1.0);
                } else {
                    int a = m[x], b = m[y];
                    double t = dfs(e, a, b);
                    res.push_back(t);
                }
            }
        }
        return res;

    }