《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 = 0): 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 = 0; i < n; ++i) {
        scanf("%d", &label);
        nodes[i] = new GraphNode(label);
    }
    
    scanf("%d", &ne);
    for (i = 0; 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 = 0; 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) == 1 && n > 0) {
        constructGraph(n, nodes);
        while (scanf("%d%d", &idx1, &idx2) == 2 && (idx1 >= 0 && idx2 >= 0)) {
            if (idx1 >= n || idx2 >= n) {
                continue;
            }

            if (hasPath(nodes[idx1], nodes[idx2], checked, path)) {
                printf("Yes
");
                printf("%d", path[0]->label);
                for (int i = 1; i < (int)path.size(); ++i) {
                    printf("->%d", path[i]->label);
                }
                printf("
");
            } else {
                printf("No
");
            }
            checked.clear();
            path.clear();
        }
        clearGraph(nodes);
    }
    
    return 0;
}