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

2014.07.04 19:34

简介：

　　给定一个带权图（有向无向皆可），找出每个顶点到其他所有顶点的最短距离。

描述：

　　此处介绍O(n^3)级别的Floyd算法，只需要用三层循环的简单代码就完成所有最短距离的计算。唯一需要注意的，就是三层循环里i、j、k的摆放顺序。

　　代码非常简单，所以无需多作解释了。

实现：

// A simple illustration for all pairs shortest path using Floyd Algorithm.
#include <iostream>
#include <vector>
using namespace std;

const int INFINITY = 1000000000;

void floydAlgorithm(const vector<vector<int> > &graph, 
    vector<vector<int> > &dist)
{
    int n;
    int i, j, k;
    
    n = (int)graph.size();
    dist.resize(n);
    for (i = 0; i < n; ++i) {
        dist[i].resize(n, INFINITY);
    }
    
    for (i = 0; i < n; ++i) {
        for (j = 0; j < n; ++j) {
            dist[i][j] = graph[i][j];
        }
    }
    
    for (k = 0; k < n; ++k) {
        for (i = 0; i < n; ++i) {
            for (j = 0; j < n; ++j) {
                if (dist[i][k] + dist[k][j] >= dist[i][j]) {
                    continue;
                }
                dist[i][j] = dist[i][k] + dist[k][j];
            }
        }
    }
}

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