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 }