CF774A. Hongcow Builds A Nation 题解 简单dfs+组合计数

题目链接：https://codeforces.com/problemset/problem/744/A

视频讲解：https://www.bilibili.com/video/BV1tZ1FYPELp?p=5

假设 \(c_i\) 所在的连通块大小（即包含节点个数）位 \(sz_i\)；

令 \(sum = \sum\limits_{i=1}^k sz_i\)，\(mx = \max\limits_{1 \le i \le k} sz_i\)

则可以新增的最多边数为 \(\sum\limits_{i=1}^k \frac{ sz_i \times (sz_i - 1) }{2} + \frac{ (n - sum) \times (n - sum - 1) }{2} + mx \times (n - sum) - m\)（具体分析可以参考视频讲解）

示例程序：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1010;int n, m, k, c[maxn], sz[maxn], sum, mx, ans;
bool vis[maxn];
vector<int> g[maxn];int dfs(int u) {if (vis[u])return 0;vis[u] = true;int res = 1;for (auto v : g[u])res += dfs(v);return res;
}int main() {scanf("%d%d%d", &n, &m, &k);for (int i = 1; i <= k; i++)scanf("%d", c+i);for (int i = 0; i < m; i++) {int u, v;scanf("%d%d", &u, &v);g[u].push_back(v);g[v].push_back(u);}for (int i = 1; i <= k; i++) {sz[i] = dfs(c[i]);sum += sz[i];mx = max(mx, sz[i]);ans += sz[i] * (sz[i] - 1) / 2;}ans += (n - sum) * (n - sum - 1) / 2 + mx * (n - sum) - m;printf("%d\n", ans);return 0;
}