UVa 216 Getting in Line【枚举排列】

题意：给出n个点的坐标，（2<=n<=8）,现在要使得这n个点连通，问最小的距离的和

因为n很小，所以可以直接枚举这n个数的排列，算每一个排列的距离的和，

保留下距离和最小的那个排列就可以了（这个地方和带宽那题有点像，用memcpy将整个排列保留下来）

 1 #include<iostream>  

 2 #include<cstdio>  

 3 #include<cstring> 

 4 #include <cmath> 

 5 #include<stack>

 6 #include<vector>

 7 #include<map> 

 8 #include<set>

 9 #include<queue> 

10 #include<algorithm>  

11 using namespace std;

12 

13 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)

14 

15 typedef long long LL;

16 const int INF = (1<<30)-1;

17 const int mod=1000000007;

18 const int maxn=100005;

19 

20 int p[maxn],bestp[maxn];

21 

22 struct node{

23     int x,y;

24 } a[maxn];

25 

26 double dis(int x1,int y1,int x2,int y2){

27      return sqrt((double)(x2 - x1) * (x2 - x1) +(double) (y2 - y1) * (y2 - y1));

28  }

29 

30 int main(){

31   // freopen("in.txt","r",stdin);

32 //    freopen("out.txt","w",stdout);

33     int  n;

34     double ans,minn=0;

35     int kase=0;

36     while(scanf("%d",&n)!=EOF&&n){

37         for(int i=0;i<n;i++) {

38             cin>>a[i].x>>a[i].y;

39             if(i!=0) minn+=dis(a[i].x,a[i].y,a[i-1].x,a[i-1].y);            

40         }

41         

42         for(int i=0;i<n;i++) p[i]=i; 

43         memcpy(bestp,p,sizeof(p));

44         

45         

46         while(next_permutation(p,p+n)){

47             ans=0;

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

49                     ans+=dis(a[p[i]].x,a[p[i]].y,a[p[i-1]].x,a[p[i-1]].y);                                

50 

51         //    printf("ans=%lf\n",ans);

52                 

53                 if(ans<minn){

54                     minn=ans;

55                     memcpy(bestp,p,sizeof(p));

56                 }

57         }

58         

59         printf("**********************************************************\n");

60         printf("Network #%d\n",++kase);

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

62             printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n",(int)a[bestp[i-1]].x,

63                     (int)a[bestp[i-1]].y,(int)a[bestp[i]].x,(int)a[bestp[i]].y,16.0 + dis(a[bestp[i-1]].x,a[bestp[i-1]].y,a[bestp[i]].x,a[bestp[i]].y));

64         printf("Number of feet of cable required is %.2lf.\n",minn + (n - 1) * 16.0);

65     }

66     return 0;

67 }

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