poj 2253 dijstra变形

dijstra算法的变形，定义：dist[i]为源点到点i的若干路径上的边的最大值的最小值，然后会发现可以用和dijstra一样的贪心方法，当前dist最小的以后都不会再被更新。

 1 #include <iostream>

 2 #include <cstring>

 3 #include <cstdio>

 4 #include <cmath>

 5 using namespace std;

 6 

 7 const int INF = 9999999;

 8 const int N = 201;

 9 int head[N];

10 int dist[N];

11 bool visit[N];

12 int n, e;

13 

14 struct Node 

15 {

16     int x, y;

17 } node[N];

18 

19 struct Edge 

20 {

21     int v, next, w;

22 } edge[N * N];

23 

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

25 {

26     edge[e].v = v;

27     edge[e].w = w;

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

29     head[u] = e++;

30 }

31 

32 int dis( int i, int j )

33 {

34     return ( node[i].x - node[j].x ) * ( node[i].x - node[j].x ) + 

35            ( node[i].y - node[j].y ) * ( node[i].y - node[j].y );

36 }

37 

38 void dij( int s )

39 {

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

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

42     {

43         dist[i] = INF;

44     }

45     dist[s] = 0;

46     for ( int i = 1; i <= n; i++ )

47     {

48         int u = 0;

49         for ( int j = 1; j <= n; j++ )

50         {

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

52             {

53                 u = j;

54             }

55         }

56         visit[u] = true;

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

58         {

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

60             if ( visit[v] ) continue;

61             dist[v] = min( dist[v], max( dist[u], w ) );

62         }

63     }

64 }

65 

66 int main ()

67 {

68     int _case = 1;

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

70     {

71         e = 0;

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

73         for ( int i = 1; i <= n; i++ )

74         {

75             scanf("%d%d", &node[i].x, &node[i].y);

76             for ( int j = i - 1; j > 0; j-- )

77             {

78                 int tmp = dis( i, j );

79                 addEdge( i, j, tmp );

80                 addEdge( j, i, tmp );

81             }

82         }

83         dij(1);

84         double ans = sqrt( dist[2] * 1.0 );

85         printf("Scenario #%d\nFrog Distance = %.3lf\n\n", _case++, ans);

86     }

87     return 0;

88 }