Codeforces Round #379 (Div. 2) E. Anton and Tree

题意：给你一棵树， 每个点要么是黑色要么是白色， 有一种操作是将同一个颜色的连通块变成相反的颜色，问你最少变换几次， 整颗树变成一种颜色。

思路： 缩点， 加求树的直径， 答案为树的直径除二向上取整。

 1 #include<bits/stdc++.h>
 2 #define LL long long
 3 #define mk make_pair
 4 using namespace std;
 5
 6 const int N = 5e5 + 7;
 7 const int inf = 0x3f3f3f3f;
 8 const int INF = 0x3f3f3f3f3f3f3f3f;
 9
10 int n, a[N], cnt[N], tot, vis[N], depth[N], id, ans;
11 vector<int> edge1[N], edge2[N];
12
13 void dfs(int u, int pre, int op, int f) {
14     vis[u] = f;
15     for(int v : edge1[u]) {
16         if(v == pre || a[v] != op)
17             continue;
18         dfs(v, u, op, f);
19     }
20 }
21
22 void dfs2(int u, int pre, int deep) {
23     depth[u] = deep;
24     if(depth[u] > depth[id])
25         id = u;
26     for(int v : edge2[u]) {
27         if(v == pre) continue;
28         dfs2(v, u, deep + 1);
29     }
30 }
31
32 void dfs3(int u, int pre, int deep) {
33     ans = max(ans, deep);
34     for(int v : edge2[u]) {
35         if(v == pre) continue;
36         dfs3(v, u, deep + 1);
37     }
38 }
39 int main() {
40     scanf("%d", &n);
41     for(int i = 1; i <= n; i++) {
42         scanf("%d", &a[i]);
43     }
44
45     for(int i = 1; i < n; i++) {
46         int u, v; scanf("%d%d", &u, &v);
47         edge1[u].push_back(v);
48         edge1[v].push_back(u);
49     }
50
51     for(int i = 1; i <= n; i++) {
52         if(vis[i]) continue;
53         dfs(i, 0, a[i], ++tot);
54     }
55
56     for(int u = 1; u <= n; u++) {
57         for(int v : edge1[u]) {
58             if(u > v || vis[u] == vis[v])
59                 continue;
60             edge2[vis[u]].push_back(vis[v]);
61             edge2[vis[v]].push_back(vis[u]);
62         }
63     }
64
65     dfs2(1, 0, 0);
66     dfs3(id, 0, 0);
67     if(ans & 1) ans++;
68     printf("%d\n", ans / 2);
69     return 0;
70 }
71 /*
72 */

原文地址：https://www.cnblogs.com/CJLHY/p/8873877.html