Educational Codeforces Round 78 (Rated for Div. 2) D. Segment Tree

Posted 2021-03-21 ydddd

链接：

https://codeforces.com/contest/1278/problem/D

题意：

As the name of the task implies, you are asked to do some work with segments and trees.

Recall that a tree is a connected undirected graph such that there is exactly one simple path between every pair of its vertices.

You are given n segments [l1,r1],[l2,r2],…,[ln,rn], li<ri for every i. It is guaranteed that all segments‘ endpoints are integers, and all endpoints are unique — there is no pair of segments such that they start in the same point, end in the same point or one starts in the same point the other one ends.

Let‘s generate a graph with n vertices from these segments. Vertices v and u are connected by an edge if and only if segments [lv,rv] and [lu,ru] intersect and neither of it lies fully inside the other one.

For example, pairs ([1,3],[2,4]) and ([5,10],[3,7]) will induce the edges but pairs ([1,2],[3,4]) and ([5,7],[3,10]) will not.

Determine if the resulting graph is a tree or not.

思路：

并差集维护关系，set查找所有能加入的点，因为最多就n个，所以不满足的条件中途就退出了，不会导致超时。

代码：

#include<bits/stdc++.h>
using namespace std;

const int MAXN = 1e6+10;

map<int, int> Mp;
int F[MAXN];
pair<int, int> Pa[MAXN];
int n;

int GetF(int x)
{
    return (x == F[x]) ? x : F[x] = GetF(F[x]);
}

int main()
{
    scanf("%d", &n);
    for (int i = 1;i <= n;i++)
        F[i] = i;
    for (int i = 1;i <= n;i++)
        cin >> Pa[i].first >> Pa[i].second;
    sort(Pa+1, Pa+1+n);
    int cnt = 0;
    for (int i = 1;i <= n;i++)
    {
        auto it = Mp.lower_bound(Pa[i].first);
        while(it != Mp.end() && it->first < Pa[i].second)
        {
            int tl = GetF(i);
            int tr = GetF(it->second);
            if (tl == tr || cnt >= n)
            {
                cout << "NO
";
                return 0;
            }
            else
            {
                cnt++;
                F[tl] = tr;
            }
            ++it;
        }
        Mp[Pa[i].second] = i;
    }
    if (cnt != n-1)
        cout << "NO
";
    else
        cout << "YES
";

    return 0;
}