《数据结构与算法分析：C语言描述》复习——第九章“图论”——无权值的最短路径问题

2014.07.04 18:24

简介：

　　给定一个有向图，你可以认为每条边长度都是1（所以叫无权值）。下面的算法可以求出从特定的起点到终点的最短路径长度。

描述：

　　从起点出发，根据当前顶点出发的边进行广度优先搜索，直至找到终点即可。如果搜索结束了仍然没有找到终点，那么起点无法到达终点。

实现：

// A simple illustration for unweighted shortest path. Graph represented by adjacency matrix.
#include <iostream>
#include <queue>
#include <vector>
using namespace std;

void unweightedShortestPath(const vector<vector<bool> > &graph, 
    vector<int> &dist, vector<bool> &reachable)
{
    // The minimal distances from 0th vertex to others.
    int n;
    int i, j;
    
    n = (int)graph.size();
    dist.resize(n);
    reachable.resize(n);

    if (n < 1) {
        return;
    }
    
    for (i = 0; i < n; ++i) {
        reachable[i] = false;
    }
    
    queue<int> q;
    
    dist[0] = 0;
    reachable[0] = true;
    
    q.push(0);
    while (!q.empty()) {
        i = q.front();
        q.pop();
        for (j = 0; j < n; ++j) {
            if (!graph[i][j] || reachable[j]) {
                continue;
            }
            dist[j] = dist[i] + 1;
            reachable[j] = true;
            q.push(j);
        }
    }
}

int main()
{
    vector<vector<bool> > graph;
    vector<int> dist;
    vector<bool> reachable;
    int n;
    int nk;
    int i, j;
    int tmp;
    
    while (cin >> n && n > 0) {
        graph.resize(n);
        for (i = 0; i < n; ++i) {
            graph[i].resize(n, false);
        }
        
        for (i = 0; i < n; ++i) {
            cin >> nk;
            for (j = 0; j < nk; ++j) {
                cin >> tmp;
                graph[i][tmp] = true;
            }
        }
        
        unweightedShortestPath(graph, dist, reachable);
        
        for (i = 0; i < n; ++i) {
            cout << i << ": ";
            if (reachable[i]) {
                cout << dist[i] << endl;
            } else {
                cout << "Unreachable" << endl;
            }
        }
        
        for (i = 0; i < n; ++i) {
            graph[i].clear();
        }
        graph.clear();
    }
    
    return 0;
}