《Cracking the Coding Interview》——第4章：树和图——题目2

2014-03-19 03:32

题目：给定一个有向图，判断其中两点是否联通。

解法：DFS搜索解决，如果是无向图的话，就可以用并查集高效解决问题了。

代码：

 // 4.2 Write a program to check if there exists a path between two nodes in a directed graph.
 #include <algorithm>
 #include <cstdio>
 #include <unordered_map>
 #include <unordered_set>
 #include <vector>
 using namespace std;

 struct  GraphNode {
     int label;
     unordered_set<GraphNode *> neighbors;

     GraphNode(int _label = ): label(_label) {};
 };

 bool hasPath(GraphNode *n1, GraphNode *n2, unordered_set<GraphNode *> &checked, vector<GraphNode *> &path)
 {
     if (n1 == nullptr || n2 == nullptr) {
         return false;
     }

     checked.insert(n1);
     path.push_back(n1);

     if (n1 == n2) {
         return true;
     }

     unordered_set<GraphNode *>::iterator it;
     for (it = n1->neighbors.begin(); it != n1->neighbors.end(); ++it) {
         if (checked.find(*it) == checked.end()) {
             if (hasPath(*it, n2, checked, path)) {
                 return true;
             }
         }
     }
     checked.erase(n1);
     path.pop_back();
     return false;
 }

 void constructGraph(int n, vector<GraphNode *> &nodes)
 {
     int i;
     int ne, x, y;
     int label;

     nodes.resize(n);
     for (i = ; i < n; ++i) {
         scanf("%d", &label);
         nodes[i] = new GraphNode(label);
     }

     scanf("%d", &ne);
     for (i = ; i < ne; ++i) {
         scanf("%d%d", &x, &y);
         nodes[x]->neighbors.insert(nodes[y]);
     }
 }

 void clearGraph(vector<GraphNode *> &nodes)
 {
     int n = (int)nodes.size();
     int i;

     for (i = ; i < n; ++i) {
         nodes[i]->neighbors.clear();
         delete nodes[i];
         nodes[i] = nullptr;
     }
     nodes.clear();
 }

 int main()
 {
     int n;
     vector<GraphNode *> nodes;
     vector<GraphNode *> path;
     unordered_set<GraphNode *> checked;
     int idx1, idx2;

     while (scanf("%d", &n) ==  && n > ) {
         constructGraph(n, nodes);
         while (scanf("%d%d", &idx1, &idx2) ==  && (idx1 >=  && idx2 >= )) {
             if (idx1 >= n || idx2 >= n) {
                 continue;
             }

             if (hasPath(nodes[idx1], nodes[idx2], checked, path)) {
                 printf("Yes\n");
                 printf("%d", path[]->label);
                 for (int i = ; i < (int)path.size(); ++i) {
                     printf("->%d", path[i]->label);
                 }
                 printf("\n");
             } else {
                 printf("No\n");
             }
             checked.clear();
             path.clear();
         }
         clearGraph(nodes);
     }

     return ;
 }