Educational Codeforces Round 36 (Rated for Div. 2)

Posted 2020-10-21 p0ny^v^

Educational Codeforces Round 36 (Rated for Div. 2)

F. Imbalance Value of a Tree

You are given a tree T consisting of n vertices. A number is written on each vertex; the number written on vertex i is ai. Let‘s denote the function I(x,?y) as the difference between maximum and minimum value of ai on a simple path connecting vertices x and y.

Your task is to calculate .

算出最大值最小值，然后减一下就是答案了。

如何计算最大值？

把点按点权升序排序，从小到大处理，从而保证当前处理的点的点权是最大的。当前点的点权乘上路径的条数就是这个点对答案的贡献了。注意要避免重复计算。具体细节看代码。

 1 #include<cstdio>
 2 #include<vector>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<algorithm>
 6 #define fir first
 7 #define sec second
 8 #define pb push_back
 9 using namespace std;
10 inline char nc() {
11     static char b[1<<17],*s=b,*t=b;
12     return s==t&&(t=(s=b)+fread(b,1,1<<17,stdin),s==t)?-1:*s++;
13 }
14 inline void read(int &x) {
15     char b = nc(); x = 0;
16     for (; !isdigit(b); b = nc());
17     for (; isdigit(b); b = nc()) x = x * 10 + b - ‘0‘;
18 }
19 typedef long long ll;
20 const int N = 1000005;
21 int n, w[N];
22 pair < int , int > a[N];
23 int sz[N], fa[N], vis[N];
24 vector < int > g[N];
25 int find(int x) {return fa[x] == x ? x : fa[x] = find(fa[x]);}
26 void merge(int a, int b) {
27     if ((a = find(a)) != (b = find(b))) {
28         if (sz[a] < sz[b]) swap(a, b);
29         fa[b] = a; sz[a] += sz[b];
30     }
31 }
32 int main() {
33     read(n); ll mx = 0, mn = 0;
34     for (int i = 1; i <= n; ++i)
35         read(w[i]), a[i] = make_pair(w[i], i);
36     for (int u, v, i = 1; i < n; ++i)
37         read(u), read(v), g[u].pb(v), g[v].pb(u);
38     sort(a + 1, a + 1 + n, less < pair < int , int > >());
39     for (int i = 1; i <= n; ++i)
40         fa[i] = i, sz[i] = 1;
41     for (int i = 1; i <= n; ++i) {
42         int u = a[i].sec; ll cnt = sz[find(u)], tcnt;
43         for (int v, j = 0; j < g[u].size(); ++j) {
44             if (w[v=g[u][j]] <= w[u] && find(u) != find(v)) {
45                 v = find(v);
46                 tcnt = sz[v];
47                 mx += cnt * tcnt * w[u];
48                 cnt += tcnt;
49                 merge(u, v);
50             }
51         }
52 //        mx += w[u];
53     }    
54     sort(a + 1, a + 1 + n, greater < pair < int , int > >());
55     for (int i = 1; i <= n; ++i)
56         fa[i] = i, sz[i] = 1;
57     for (int i = 1; i <= n; ++i) {
58         int u = a[i].sec; ll cnt = sz[find(u)], tcnt;    
59         for (int v, j = 0; j < g[u].size(); ++j) {
60             if (w[v=g[u][j]] >= w[u] && find(u) != find(v)) {
61                 v = find(v);
62                 tcnt = sz[v];
63                 mn += cnt * tcnt * w[u];
64                 cnt += tcnt;
65                 merge(u, v);
66             }
67         }
68 //        mn += w[u];
69     }    
70     printf("%lld\n", mx - mn);
71     return 0;    
72 }