UVa 1354 枚举子集 Mobile Computing

只要枚举左右两个子天平砝码的集合，我们就能算出左右两个悬挂点到根悬挂点的距离。

但是题中要求找尽量宽的天平但是不能超过房间的宽度，想不到要怎样记录结果。

参考别人代码，用了一个结构体的vector，保存每个集合合法方案的左右两端最长的距离。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <vector>
 5 #include <map>
 6 #include <cmath>
 7 #define MP make_pair
 8 #define Ft first
 9 #define Sd second
10 using namespace std;
11
12 typedef pair<double, double> PDD;
13
14 const int maxn = 10;
15 const int maxs = 1000;
16
17 vector<PDD> tree[maxs];
18
19 int n;
20 double r;
21 double a[maxn], w[maxs];
22 bool vis[maxs];
23
24 int bitcount(int x)
25 {
26     int ans = 0;
27     while(x) { ans += (x & 1); x >>= 1; }
28     return ans;
29 }
30
31 void dfs(int S)
32 {
33     if(vis[S]) return ;
34     vis[S] = true;
35     if(bitcount(S) == 1) { tree[S].push_back(MP(0, 0)); return ; }
36
37     PDD t = MP(0, 0);
38     for(int s1 = (S-1)&S; s1; s1 = (s1-1)&S)
39     {
40         int s2 = S ^ s1;
41         dfs(s1); dfs(s2);
42         double x1 = w[s2] / w[S], x2 = w[s1] / w[S];
43         for(int i = 0; i < tree[s1].size(); i++)
44             for(int j = 0; j < tree[s2].size(); j++)
45             {
46                 t.Ft = max(x1 + tree[s1][i].Ft, tree[s2][j].Ft - x2);
47                 t.Sd = max(x2 + tree[s2][j].Sd, tree[s1][i].Sd - x1);
48                 if(t.Ft + t.Sd < r) tree[S].push_back(t);
49             }
50     }
51 }
52
53 int main()
54 {
55     int T; scanf("%d", &T);
56     while(T--)
57     {
58         scanf("%lf%d", &r, &n);
59         for(int i = 0; i < n; i++) scanf("%lf", a + i);
60         int all = (1 << n) - 1;
61
62         for(int i = 0; i <= all; i++)
63         {
64             w[i] = 0;
65             tree[i].clear();
66             for(int j = 0; j < n; j++) if(i & (1 << j))
67                 w[i] += a[j];
68         }
69
70         memset(vis, false, sizeof(vis));
71         dfs(all);
72         double ans = -1;
73         for(int i = 0; i < tree[all].size(); i++) ans = max(ans, tree[all][i].Ft + tree[all][i].Sd);
74         if(ans < 0) puts("-1");
75         else printf("%.9f\n", ans);
76     }
77
78     return 0;
79 }

代码君