Codeforces Round #738 (Div. 2)_D2. Mocha and Diana (Hard Version)_贪心并查集

D2. Mocha and Diana (Hard Version)

题目传送门：

题目传送门

题面：

题目大意：

相比easy version只改变了数据范围。

思路：

贪心+并查集。
先都看看能不能和1连，再把散点看看能不能一一配对。

代码：

#include<bits/stdc++.h>

using namespace std;
const int maxn = 1e5 + 10;


int p1[maxn], p2[maxn];

void init(int fa[], int n) {
    for (int i = 1; i <= n; ++i)
        fa[i] = i;
}

int find1(int x) {
    if (p1[x] != x) p1[x] = find1(p1[x]);
    return p1[x];
}

void merge1(int x, int y) {
    x = find1(x);
    y = find1(y);
    if (x > y)swap(x, y);
    p1[y] = x;
}

int find2(int x) {
    if (p2[x] != x) p2[x] = find2(p2[x]);
    return p2[x];
}

void merge2(int x, int y) {
    x = find2(x);
    y = find2(y);
    if (x > y)swap(x, y);
    p2[y] = x;
}


int main() {
    int n, m1, m2;
    scanf("%d%d%d", &n, &m1, &m2);
    int u, v;
    init(p1, n);
    init(p2, n);
    for (int i = 0; i < m1; i++) {
        scanf("%d%d", &u, &v);
        merge1(u, v);
    }
    for (int i = 0; i < m2; i++) {
        scanf("%d%d", &u, &v);
        merge2(u, v);
    }
    vector<pair<int, int>> ans;
    vector<int> v1, v2;
    for (int i = 2; i <= n; i++) {
        if (find1(i) != 1 && find2(i) != 1) {
            ans.push_back({1, i});
            merge1(1, i);
            merge2(1, i);
        } else if (find1(i) != 1) {
            v1.push_back(i);
        } else if (find2(i) != 1) {
            v2.push_back(i);
        }
    }
    while (!v1.empty() && !v2.empty()) {
        if (find1(v1.back()) == 1 && find2(v1.back()) == 1) {
            //此时仅剩下一张图中与1连或者两张图都和1连的情况
            //此时属于后者，如果存在有点能连，那么那个点要在两张图中都未连上1
            //已经没有这样的点了
            //直接寄
            v1.pop_back();
            continue;
        }
        if (find1(v2.back()) == 1 && find2(v2.back()) == 1) {
            //同上
            v2.pop_back();
            continue;
        }
        ans.push_back({v1.back(), v2.back()});
        //1,2中剩下的到不了的散点之间互相连起来
        merge1(v1.back(), v2.back());
        merge2(v1.back(), v2.back());
        v1.pop_back();
        v2.pop_back();
    }
    cout << ans.size() << '\n';
    for (auto it : ans) {
        cout << it.first << ' ' << it.second << '\n';
    }

}