USACO2.43Cow Tours（floyd+dfs)

dfs用来求这个节点属于哪一块 用floyd求出各点间的最短距离 再求出每个节点可达到的最远距离MAX[i]以及这两个块中的最长距离a,b. 枚举当两个节点不在一块中 dis = max{(dis[i][j]+MAX[i]+MAX[j] ),a,b),在求出dis中最小的。

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 1 /*

 2      ID: shangca2

 3      LANG: C++

 4      TASK: cowtour

 5  */

 6 #include <iostream>

 7 #include<cstdio>

 8 #include<cstring>

 9 #include<algorithm>

10 #include<stdlib.h>

11 #include<cmath>

12 using namespace std;

13 #define INF 0x3f3f3f

14 struct node

15 {

16     int x,y;

17 }no[200];

18 double w[200][200],m[200],a[5];

19 int n,vis[200];

20 void dfs(int x,int d)

21 {

22     int i;

23     for(i = 1; i <= n ; i++)

24         if(!vis[i]&&w[x][i]!=INF)

25         {

26             vis[i] = d;

27             dfs(i,d);

28         }

29 }

30 int main()

31 {

32     freopen("cowtour.in","r",stdin);

33     freopen("cowtour.out","w",stdout);

34     int i,j,k;

35     char c;

36     cin>>n;

37     for(i = 1; i <= n ; i++)

38     cin>>no[i].x>>no[i].y;

39     for(i = 1; i <= n ; i++)

40     {

41         for(j = 1; j <= n ; j++)

42         {

43             cin>>c;

44             if(c=='1')

45             w[i][j] = 1.0*sqrt((no[i].x-no[j].x)*(no[i].x-no[j].x)+(no[i].y-no[j].y)*(no[i].y-no[j].y));

46             else

47             w[i][j] = INF;

48         }

49     }

50     int g = 1;

51     for(i = 1; i <= n ; i++)

52         if(!vis[i])

53         {

54             vis[i] = g;

55             dfs(i,g++);

56         }

57     for(i = 1; i <= n ; i++)

58     w[i][i] = 0;

59     for(i = 1; i <= n ; i++)

60         for(j = 1; j <= n ; j++)

61             for(k = 1; k <= n ; k++)

62             {

63                 if(w[j][k]>w[j][i]+w[i][k])

64                 w[j][k]=w[j][i]+w[i][k];

65             }

66     for(i = 1; i <= n ; i++)

67         for(j = 1; j <= n ;j++)

68         {

69             if(w[i][j]<INF)

70             a[vis[i]] = max(a[vis[i]],w[i][j]);

71         }

72     for(i = 1; i <= n ; i++)

73     {

74         m[i] = 0;

75         for(j = 1; j <= n ; j++)

76         if(w[i][j]<INF)

77         m[i] = max(m[i],w[i][j]);

78     }

79     double mi=INF,mm = INF;

80 

81     for(i = 1; i <= n ; i++)

82     {

83         for(j = 1; j <= n ; j++)

84         {

85             if(vis[i]!=vis[j])

86             {

87                 mi = m[i]+m[j]+1.0*sqrt((no[i].x-no[j].x)*(no[i].x-no[j].x)+(no[i].y-no[j].y)*(no[i].y-no[j].y));

88                 mi = max(mi,a[1]);

89                 mi = max(mi,a[2]);

90             }

91             if(mi<mm)

92             mm = mi;

93         }

94     }

95     printf("%.6lf\n",mm);

96     return 0;

97 }