题目传送门

　　传送门I

　　传送门II

题目大意

　　给定一棵带边权有根树，修改一条边的边权的代价是修改前和修改后的值的绝对值之差。不能将一条边的边权改为负数。问使得根节点到所有叶节点的距离相等的最小代价。

 1 /**
 2  * bzoj
 3  * Problem#4585
 4  * Accepted
 5  * Time: 8736ms
 6  * Memory: 18872k
 7  */
 8 #include <iostream>
 9 #include <cassert>
10 #include <cstdlib>
11 #include <cstring>
12 #include <cstdio>
13 #ifndef WIN32
14 #define Auto "%lld"
15 #else
16 #define Auto "%I64d"
17 #endif
18 using namespace std;
19 typedef bool boolean;
20
21 #define ll long long
22 const int N = 6e5 + 5;
23
24 typedef class SkewNode {
25     public:
26         ll val;
27         SkewNode *l, *r;
28
29         SkewNode()    {    }
30 }SkewNode;
31
32 SkewNode pool[N];
33 SkewNode* top = pool;
34
35 SkewNode* newnode(int val) {
36     top->val = val;
37     return top++;
38 }
39
40 SkewNode* merge(SkewNode* a, SkewNode* b) {
41     if (!a || !b)    return (a) ? (a) : (b);
42     if (a->val < b->val)    swap(a, b);
43     a->r = merge(a->r, b);
44     swap(a->l, a->r);
45     return a;
46 }
47
48 int n, m;
49 int *fa, *cs, *ss, *ks;
50 ll *bs;
51 SkewNode** rs;
52
53 inline void init() {
54     scanf("%d%d", &n, &m);
55     n += m;
56     bs = new ll[(n + 1)];
57     fa = new int[(n + 1)];
58     cs = new int[(n + 1)];
59     ss = new int[(n + 1)];
60     ks = new int[(n + 1)];
61     rs = new SkewNode*[(n + 1)];
62     memset(bs, 0, sizeof(ll) * (n + 1));
63     memset(ks, 0, sizeof(int) * (n + 1));
64     memset(ss, 0, sizeof(int) * (n + 1));
65     memset(rs, 0, sizeof(SkewNode*) * (n + 1));
66     for (int i = 2; i <= n; i++)
67         scanf("%d%d", fa + i, cs + i);
68 }
69
70 inline void solve() {
71     for (int i = n - m + 1; i <= n; i++) {
72         int f = fa[i];
73         ss[f] += 2, ks[f] += 1, bs[f] += cs[i];
74         rs[f] = merge(rs[f], newnode(cs[i]));
75         rs[f] = merge(rs[f], newnode(cs[i]));
76     }
77     for (int i = n - m; i > 1; i--) {
78         while (ss[i] > ks[i] + 1)    ss[i]--, rs[i] = merge(rs[i]->l, rs[i]->r);
79         SkewNode *a = rs[i], *b = rs[i] = merge(rs[i]->l, rs[i]->r);
80         bs[i] += cs[i], a->val += cs[i], b->val += cs[i], a->l = a->r = NULL;
81         rs[i] = merge(rs[i], a);
82         int f = fa[i];
83         bs[f] += bs[i], ks[f] += ks[i], ss[f] += ss[i];
84         rs[f] = merge(rs[f], rs[i]);
85     }
86     while (ss[1] > ks[1])    ss[1]--, rs[1] = merge(rs[1]->l, rs[1]->r);
87     ll res = bs[1];
88     while (rs[1])
89         res -= rs[1]->val, rs[1] = merge(rs[1]->l, rs[1]->r);
90     printf(Auto"\n", res);
91 }
92
93 int main() {
94     init();
95     solve();
96     return 0;
97 }