题解 | #小欧皇#

每个连通块对答案的贡献是固定的，如果连通块内有 n 个点，那么答案就是 n*(n-1)/2 ，所以我们可以用并查集维护出所有的连通块，然后维护出初始的答案，然后枚举剩下的所有为 0 的点变成 1 以后的情况，他会连接所有的相邻的 1 的连通块

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int INF = 1e18;
const int N = 1e5 + 5;
int __t = 1, n, u, v, sum, m, f[N], w[N], vis[N];
string s;
vector<int> g[N];

int find(int x) {
    return x == f[x] ? x : f[x] = find(f[x]);
}

void mg(int x, int y) {
    x = find(x), y = find(y);
    if (x != y) {
        f[x] = y;
        w[y] += w[x];
    }
}

void solve() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        f[i] = i, w[i] = 1;
    cin >> s;
    s = " " + s;
    for (int i = 0; i < m; i++) {
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
        if (s[u] == '1' && s[v] == '1')
            mg(u, v);
    }
    for (int i = 1; i <= n; i++) {
        if (s[i] == '1') {
            int fa = find(i);
            if (!vis[fa]) {
                vis[fa] = 1;
                sum += w[fa] * (w[fa] - 1) / 2;
            }
        }
    }
    int ansi = 0, anssum = 0;
    for (int i = 1; i <= n; i++) {
        if (s[i] == '0') {
            map<int, int> mp;
            int ssum = sum, num = 1;
            for (int v : g[i]) {
                if (s[v] == '1') {
                    int x = find(v);
                    if (mp[x])
                        continue;
                    mp[x] = 1;
                    ssum -= w[x] * (w[x] - 1) / 2;
                    num += w[x];
                }
            }
            ssum += num * (num - 1) / 2;
            if (ssum > anssum) {
                anssum = ssum;
                ansi = i;
            }
        }
    }
    cout << ansi << " " << anssum << "\n";
}
int32_t main() {
#ifdef ONLINE_JUDGE
    ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    // cin >> __t;
    while (__t--)
        solve();
    return 0;
}