图的广度遍历

题目描述

现有一些格式为A / B = k的等式，A和B均为字符串变量，k为一个实数（浮点数）。如果能够找到答案，则返回答案。若答案不存在，则返回-1.0。

样例
给出等式a / b = 2.0, b / c = 3.0。
问题为a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?。

返回[6.0, 0.5, -1.0, 1.0, -1.0 ]。

代码如下：

#include <iostream>#include <string>#include <vector>#include <map>#include <utility>#include <queue>using namespace std;#define maxn 128struct Node{double v;int index;Node(int _index, double _v) :v(_v), index(_index){};};queue<Node> q;map<string, int> mp;pair<string, string> a[] = { make_pair("a", "b"), make_pair("b", "c"), make_pair("d", "c") };double g[maxn][maxn];vector<Node> eg[maxn];double values[] = { 2.0, 3.0 ,2.0};//得数pair<string, string> queries[] = { make_pair("a", "b"), make_pair("c", "a") };double FindSolve(pair<string, string> p){int s = mp[p.first];int t = mp[p.second];q.push(Node(s,1));while (!q.empty()){Node cur = q.front(); q.pop();if (cur.index == t){return cur.v;}for (int i = 0; i < eg[cur.index].size(); i++){int index = eg[cur.index][i].index;double v = eg[cur.index][i].v;q.push(Node(index,cur.v*v));}}return  -1.0;}int main(){int cnt = 1;//给字符串编号for (int i = 0; i < 3; i++){if (mp[a[i].first] == NULL){mp[a[i].first] = cnt++;}if (mp[a[i].second] == NULL){mp[a[i].second] = cnt++;}eg[mp[a[i].first]].push_back(Node(mp[a[i].second],values[i]));eg[mp[a[i].second]].push_back(Node(mp[a[i].first], 1.0/values[i]));//g[mp[a[i].first]][mp[a[i].second]] = values[i];}for (int i = 0; i < 2; i++){printf("%lf\n", FindSolve(queries[i]));}return 0;}