《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：

 1 // A simple illustration for all pairs shortest path using Floyd Algorithm.

 2 #include <iostream>

 3 #include <vector>

 4 using namespace std;

 5 

 6 const int INFINITY = 1000000000;

 7 

 8 void floydAlgorithm(const vector<vector<int> > &graph, 

 9     vector<vector<int> > &dist)

10 {

11     int n;

12     int i, j, k;

13     

14     n = (int)graph.size();

15     dist.resize(n);

16     for (i = 0; i < n; ++i) {

17         dist[i].resize(n, INFINITY);

18     }

19     

20     for (i = 0; i < n; ++i) {

21         for (j = 0; j < n; ++j) {

22             dist[i][j] = graph[i][j];

23         }

24     }

25     

26     for (k = 0; k < n; ++k) {

27         for (i = 0; i < n; ++i) {

28             for (j = 0; j < n; ++j) {

29                 if (dist[i][k] + dist[k][j] >= dist[i][j]) {

30                     continue;

31                 }

32                 dist[i][j] = dist[i][k] + dist[k][j];

33             }

34         }

35     }

36 }

37 

38 int main()

39 {

40     vector<vector<int> > graph;

41     vector<vector<int> > dist;

42     int n;

43     int nk;

44     int i, j;

45     int tmp, tmp_dist;

46     

47     while (cin >> n && n > 0) {

48         graph.resize(n);

49         for (i = 0; i < n; ++i) {

50             graph[i].resize(n, INFINITY);

51         }

52         

53         for (i = 0; i < n; ++i) {

54             cin >> nk;

55             for (j = 0; j < nk; ++j) {

56                 cin >> tmp >> tmp_dist;

57                 graph[i][tmp] = tmp_dist;

58             }

59         }

60         

61         floydAlgorithm(graph, dist);

62         

63         for (i = 0; i < n; ++i) {

64             for (j = 0; j < n; ++j) {

65                 cout << "[" << i << "][" << j << "]: ";

66                 if (dist[i][j] < INFINITY) {

67                     cout << dist[i][j] << endl;

68                 } else {

69                     cout << "Unreachable" << endl;

70                 }

71             }

72         }

73         

74         for (i = 0; i < n; ++i) {

75             graph[i].clear();

76         }

77         graph.clear();

78         

79         for (i = 0; i < n; ++i) {

80             dist[i].clear();

81         }

82         dist.clear();

83     }

84     

85     return 0;

86 }