BZOJ4033 [HAOI2015]T1

令$f[p][i]$表示以$p$为根的子树内，选了$i$个黑点，剩下的都是白点的这个子树内贡献的答案

如果$p$的子树都算出来了，只要计算$p$与$fa[p]$之间的边对答案的贡献就好了，贡献是$dis * (i * (sz - i) + (k - i) * (n - k - (sz - i)))$

于是树形动规一下就好了

 

 1 /**************************************************************
 2     Problem: 4033
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:308 ms
 7     Memory:32300 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13  
14 using namespace std;
15 typedef long long ll;
16  
17 const int N = 2e3 + 5;
18  
19 struct edge {
20     int next, to, v;
21     edge(int _n = 0, int _t = 0, int _v = 0) : next(_n), to(_t), v(_v) {}
22 } e[N << 1];
23  
24 int n, k;
25 int first[N], tot;
26 int sz[N];
27 ll f[N][N], tmp[N];
28  
29 inline void Add_Edges(int x, int y, int z) {
30     e[++tot] = edge(first[x], y, z), first[x] = tot;
31     e[++tot] = edge(first[y], x, z), first[y] = tot;
32 }
33  
34 inline ll calc(int x, int y) {
35     return 1ll * y * (x - y);
36 }
37  
38 #define y e[x].to
39 inline void dfs(int p, int fa, int w) {
40     int x, i, j;
41     sz[p] = 1;
42     for (x = first[p]; x; x = e[x].next)
43         if (y != fa) {
44             dfs(y, p, e[x].v);
45             memcpy(tmp, f[p], sizeof(tmp));
46             for (i = min(sz[p], k); ~i; --i)
47                 for (j = min(sz[y], k - i); ~j; --j)
48                     tmp[i + j] = max(tmp[i + j], f[p][i] + f[y][j]);
49             memcpy(f[p], tmp, sizeof(tmp));
50             sz[p] += sz[y];
51         }
52     for (i = min(sz[p], k); ~i; --i)
53         f[p][i] += (calc(k, i) + calc(n - k, sz[p] - i)) * w;
54 }
55 #undef y
56  
57 int main() {
58     int i, x, y, z;
59     scanf("%d%d", &n, &k);
60     for (i = 1; i < n; ++i) {
61         scanf("%d%d%d", &x, &y, &z);
62         Add_Edges(x, y, z);
63     }
64     dfs(1, 0, 0);
65     printf("%lld\n", f[1][k]);
66     return 0;
67 }
View Code

 

posted on 2015-05-04 22:08  Xs酱~  阅读(385)  评论(0)  编辑  收藏  举报