AtCoder Beginner Contest 238

E - Range Sums

https://atcoder.jp/contests/abc238/tasks/abc238_e

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $500$ points

Problem Statement

Takahashi has a secret integer sequence $a$. You know that the length of $a$ is $N$.

You want to guess the contents of $a$. He has promised to give you the following $Q$ additional pieces of information.

  • The $i$-th information: the value $a_{l_i}+a_{l_i+1}+\cdots+a_{r_i}$.

Is it possible to determine the sum of all elements in $a$, $a_1+a_2+\cdots+a_N$, if the $Q$ pieces of promised information are given?

Constraints

  • $1 \leq N \leq 2 \times 10^5$
  • $1 \leq Q \leq \min(2 \times 10^5,\frac{N(N+1)}{2})$
  • $1 \leq l_i \leq r_i \leq N$
  • $(l_i,r_i) \neq (l_j,r_j)\ (i \neq j)$
  • All values in input are integers.

Input

Input is given from Standard Input in the following format:

$N$ $Q$
$l_1$ $r_1$
$l_2$ $r_2$
$\hspace{0.4cm}\vdots$
$l_Q$ $r_Q$

Output

If it is possible to determine the sum of all elements in $a$, print Yes; otherwise, print No.


Sample Input 1

3 3
1 2
2 3
2 2

Sample Output 1

Yes

From the first and second information, we can find the value $a_1+a_2+a_2+a_3$. By subtracting the value of $a_2$ from it, we can determine the value $a_1+a_2+a_3$.


Sample Input 2

4 3
1 3
1 2
2 3

Sample Output 2

No

We can determine the sum of the first $3$ elements of $a$, but not the sum of all elements.


Sample Input 3

4 4
1 1
2 2
3 3
1 4

Sample Output 3

Yes

The fourth information directly gives us the sum of all elements.

我的笔记

如果能想到并查集这题非常的简单。

首先有 $N$ 个数,我们把这些数排成一排,那么就有 $N+1$ 个间隔(包括两端),然后我们把每一个 $l$ $r$ 对应到这 $N+1$ 个间隔,只要最后 $0\sim N+1$ 相连,那么就说明可以求出和。

如下图,表示的就是 1 6;2 4;2 6;5 8 这四个联通,最后 0 和 8 联通,即可以求出和。

求联通,很简单想到用并查集,因此初始化一个大小为 $N+1$ 的并查集即可。

另外,这题必须要用路径优化,否则会有两个测试点 TLE,按秩合并倒不重要,实测按秩合并效率反而略低。

源码

#include <bits/stdc++.h>
#define MAXN 200010

using namespace std;

int fa[MAXN];
inline void init(int n)
{
    for (int i = 0; i < n; i++)
        fa[i] = i;
}

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

inline void merge(int i, int j)
{
    fa[find(i)] = find(j);
}

int main(void)
{
    ios::sync_with_stdio(false);
    int N, Q;
    cin >> N >> Q;
    init(N + 1);
    while (Q--)
    {
        int l, r;
        cin >> l >> r;
        merge(l - 1, r);
    }
    if (find(0) == find(N))
    {
        cout << "Yes" << endl;
    }
    else
    {
        cout << "No" << endl;
    }
    return 0;
}

标签: 思维, 并查集

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