Codeforces Round #501 (Div. 3) 1015A Points in Segments （前缀和）

A. Points in Segments

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a set of $\mathrm{nn\; segments\; on\; the\; axis\; OxOx\; ,\; each\; segment\; has\; integer\; endpoints\; between\; 11\; and\; mm\; inclusive.\; Segments\; may\; intersect,\; overlap\; or\; even\; coincide\; with\; each\; other.\; Each\; segment\; is\; characterized\; by\; two\; integers\; lili\; and\; riri\; (1\le li\le ri\le m1\le li\le ri\le m\; )\; —\; coordinates\; of\; the\; left\; and\; of\; the\; right\; endpoints.}$

Consider all integer points between $\mathrm{11\; and\; mm\; inclusive.\; Your\; task\; is\; to\; print\; all\; such\; points\; that\; don\text{'}t\; belong\; to\; any\; segment.\; The\; point\; xx\; belongs\; to\; the\; segment\; [l;r][l;r]\; if\; and\; only\; if\; l\le x\le rl\le x\le r\; .}$

Input

The first line of the input contains two integers $\mathrm{nn\; and\; mm\; (1\le n,m\le 1001\le n,m\le 100\; )\; —\; the\; number\; of\; segments\; and\; the\; upper\; bound\; for\; coordinates.}$

The next $\mathrm{nn\; lines\; contain\; two\; integers\; each\; lili\; and\; riri\; (1\le li\le ri\le m1\le li\le ri\le m\; )\; —\; the\; endpoints\; of\; the\; ii\; -th\; segment.\; Segments\; may\; intersect,\; overlap\; or\; even\; coincide\; with\; each\; other.\; Note,\; it\; is\; possible\; that\; li=rili=ri\; ,\; i.e.\; a\; segment\; can\; degenerate\; to\; a\; point.}$

Output

In the first line print one integer $\mathrm{kk\; —\; the\; number\; of\; points\; that\; don\text{'}t\; belong\; to\; any\; segment.}$

In the second line print exactly $\mathrm{kk\; integers\; in\; any\; order\; —\; the\; points\; that\; don\text{'}t\; belong\; to\; any\; segment.\; All\; points\; you\; print\; should\; be\; distinct.}$

If there are no such points at all, print a single integer $\mathrm{00\; in\; the\; first\; line\; and\; either\; leave\; the\; second\; line\; empty\; or\; do\; not\; print\; it\; at\; all.}$

Examples

Input

Copy

3 5
2 2
1 2
5 5

Output

Copy

2
3 4 

Input

Copy

1 7
1 7

Output

Copy

0

Note

In the first example the point $\mathrm{11\; belongs\; to\; the\; second\; segment,\; the\; point\; 22\; belongs\; to\; the\; first\; and\; the\; second\; segments\; and\; the\; point\; 55\; belongs\; to\; the\; third\; segment.\; The\; points\; 33\; and\; 44\; do\; not\; belong\; to\; any\; segment.}$

In the second example all the points from $\mathrm{11\; to\; 77\; belong\; to\; the\; first\; segment.}$

 

 

 

题目大意：n个线段 [l, r], 然后1~m个点，输出哪些点不属于任何线段。

 

 

 

当然是个水题，时间复杂度O(n*m), 如果  n,m <= 10^7呢？

可以用前缀和O(n+m)解答：

给出  l,r  暴力想法直接 vis[l]~vis[r] 标记 1，但是有更好的想法。

将 vis[l]++,  vis[r+1]-- ,求前缀和的时候，虽然中间的没有标记1，但是前缀和往后延申就达到标记效果了。将vis[r+1]--, 传到r+1的时候就（-1） + 1 = 0,  也就是没有出现过了。

例如 ：  点1  2  3 4 5

给出线段 【2 4】， 【4，5】；可以看出答案为   1

点              0    1    2     3      4      5       6

[2,4]           0     0    1     0     0      -1      0

[4,5]           0     0    1     0     1       0      -1

前缀和       0     0    1     1     2       2       1（只有点1为0）

 

 

 

AC代码(原题解)：

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
#endif
    
    int n, m;
    cin >> n >> m;
    vector<int> cnt(m + 2);
    for (int i = 0; i < n; ++i) {
        int l, r;
        cin >> l >> r;
        ++cnt[l];
        --cnt[r + 1];
    }
    for (int i = 1; i <= m; ++i)
        cnt[i] += cnt[i - 1];
    vector<int> ans;
    for (int i = 1; i <= m; ++i) {
        if (cnt[i] == 0)
            ans.push_back(i);
    }
    
    cout << ans.size() << endl;
    for (auto it : ans) cout << it << " ";
    cout << endl;
    
    return 0;
}