BZOJ3754 Tree之最小方差树

Orz AutSky_JadeK他已经讲的很详细了，我只是个酱油233

（p.s. 这输出样例是错的。。。对的应该是0.7071貌似= =）

 1 /**************************************************************
 2     Problem: 3754
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:9024 ms
 7     Memory:892 kb
 8 ****************************************************************/
 9
10 #include <cstdio>
11 #include <cmath>
12 #include <algorithm>
13
14 using namespace std;
15 typedef double lf;
16 const int N = 105;
17 const int M = 2005;
18 const int Maxlen = 30005;
19
20 struct edges {
21     int x, y, v;
22     lf w;
23
24     inline bool operator < (const edges &b) const {
25         return w < b.w;
26     }
27 } e[M];
28 inline bool cmp_min(edges a, edges b) {
29     return a.v < b.v;
30 }
31 inline bool cmp_max(edges a, edges b) {
32     return a.v > b.v;
33 }
34
35
36 int n,m, Min, Max, fa[N];
37 lf ans = (lf) 1e15;
38
39 char buf[Maxlen], *c = buf;
40 int Len;
41
42 inline int read() {
43     int x = 0;
44     while (*c < ‘0‘ || ‘9‘ < *c) ++c;
45     while (‘0‘ <= *c && *c <= ‘9‘)
46         x = x * 10 + *c - ‘0‘, ++c;
47     return x;
48 }
49
50 lf Sqr(lf x) {
51     return x * x;
52 }
53
54 int find(int x) {
55     return x == fa[x] ? x : fa[x] = find(fa[x]);
56 }
57
58 void work(int sum) {
59     lf A = (lf) sum / (n - 1), res = 0;
60     int cost = 0, tot = 0, fa1, fa2, i;
61     for (i = 1; i <= n; ++i)
62         fa[i] = i;
63     for (i = 1; i <= m; ++i)
64         e[i].w = Sqr((lf) e[i].v - A);
65     sort(e + 1, e + m + 1);
66     for (i = 1; i <= m; ++i)
67         if ((fa1 = find(e[i].x)) != (fa2 = find(e[i].y))) {
68             fa[fa1] = fa2;
69             cost += e[i].v;
70             res += e[i].w;
71             if (++tot == n - 1) break;
72         }
73     if (cost == sum)
74         ans = min(ans, res);
75 }
76
77 int main() {
78     int i;
79     Len = fread(c, 1, Maxlen, stdin);
80     buf[Len] = ‘\0‘;
81     n = read(), m = read();
82     for (i = 1; i <= m; ++i)
83         e[i].x = read(), e[i].y = read(), e[i].v = read();
84     sort(e + 1, e + m + 1, cmp_min);
85     for (i = 1; i < n; ++i)
86         Min += e[i].v;
87     sort(e + 1, e + m + 1, cmp_max);
88     for (i = 1; i < n; ++i)
89         Max += e[i].v;
90     for (i = Min; i <= Max; ++i)
91         work(i);
92     printf("%.4lf\n", sqrt((lf) ans / (n - 1)));
93 }