toj 2870 理解dijstra

题意：求离一个城市第k远的城市的编号，无负权边。

思路：根据dijstra算法的特点可以知道，外层循环n次，每次求出的点就是离源点第i远的点的最短路。所以迭代k次，第k次新求出的点即为答案。

 1 #include <algorithm>

 2 #include <iostream>

 3 #include <cstring>

 4 #include <cstdio>

 5 using namespace std;

 6 

 7 const int INF = 9999999;

 8 const int N = 201;

 9 const int M = 20000;

10 int head[N];

11 int dist[N];

12 bool visit[N];

13 int n, m, k, e;

14 

15 struct Edge

16 {

17     int v, next, w;

18 } edge[M];

19 

20 void addEdge( int u, int v, int w )

21 {

22     edge[e].v = v;

23     edge[e].w = w;

24     edge[e].next = head[u];

25     head[u] = e++;

26 }

27 

28 int dij( int s )

29 {

30     memset( visit, false, sizeof(visit) );

31     for ( int i = 0; i <= n; i++ )

32     {

33         dist[i] = INF;

34     }

35     dist[s] = 0;

36     int u;

37     for ( int i = 0; i <= k; i++ )

38     {

39         u = n;

40         for ( int j = 0; j < n; j++ )

41         {

42             if ( !visit[j] && dist[j] < dist[u] )

43             {

44                 u = j;

45             }

46         }

47         visit[u] = true;

48         for ( int j = head[u]; j != -1; j = edge[j].next )

49         {

50             int v = edge[j].v, w = edge[j].w;

51             if ( !visit[v] && dist[u] + w < dist[v] )

52             {

53                 dist[v] = dist[u] + w;

54             }

55         }

56     }

57     return u;

58 }

59 

60 int main ()

61 {

62     while ( scanf("%d", &n), n )

63     {

64         scanf("%d", &m);

65         e = 0;

66         memset( head, -1, sizeof(head) );

67         while ( m-- )

68         {

69             int u, v, w;

70             scanf("%d%d%d", &u, &v, &w);

71             addEdge( u, v, w );

72             addEdge( v, u, w );

73         }

74         scanf("%d", &k);

75         printf("%d\n", dij(0));

76     }

77     return 0;

78 }