htm multiset rip namespace next msu == nec -i

C. Maximal Intersection time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output

You are given $\mathrm{nn\; segments\; on\; a\; number\; line;\; each\; endpoint\; of\; every\; segment\; has\; integer\; coordinates.\; Some\; segments\; can\; degenerate\; to\; points.\; Segments\; can\; intersect\; with\; each\; other,\; be\; nested\; in\; each\; other\; or\; even\; coincide.}$

The intersection of a sequence of segments is such a maximal set of points (not necesserily having integer coordinates) that each point lies within every segment from the sequence. If the resulting set isn‘t empty, then it always forms some continuous segment. The length of the intersection is the length of the resulting segment or

00 in case the intersection is an empty set.

For example, the intersection of segments $[1;5][1;5]\; and\; [3;10][3;10]\; is\; [3;5][3;5]\; (length\; 22),\; the\; intersection\; of\; segments\; [1;5][1;5]\; and\; [5;7][5;7]\; is\; [5;5][5;5](length\; 00)\; and\; the\; intersection\; of\; segments\; [1;5][1;5]\; and\; [6;6][6;6]\; is\; an\; empty\; set\; (length\; 00).$

Your task is to remove exactly one segment from the given sequence in such a way that the intersection of the remaining $(n?1)(n?1)segments\; has\; the\; maximal\; possible\; length.$

Input

The first line contains a single integer $\mathrm{nn\; (2\le n\le 3?1052\le n\le 3?105)\; —\; the\; number\; of\; segments\; in\; the\; sequence.}$

Each of the next $\mathrm{nn\; lines\; contains\; two\; integers\; lili\; and\; riri\; (0\le li\le ri\le 1090\le li\le ri\le 109)\; —\; the\; description\; of\; the\; ii-th\; segment.}$

Output

Print a single integer — the maximal possible length of the intersection of $(n?1)(n?1)\; remaining\; segments\; after\; you\; remove\; exactly\; one\; segment\; from\; the\; sequence.$

Examples input Copy

4
1 3
2 6
0 4
3 3

output Copy

1

input Copy

5
2 6
1 3
0 4
1 20
0 4

output Copy

2

input Copy

3
4 5
1 2
9 20

output Copy

0

input Copy

2
3 10
1 5

output Copy

7

Note

In the first example you should remove the segment $[3;3][3;3],\; the\; intersection\; will\; become\; [2;3][2;3]\; (length\; 11).\; Removing\; any\; other\; segment\; will\; result\; in\; the\; intersection\; [3;3][3;3]\; (length\; 00).$

In the second example you should remove the segment $[1;3][1;3]\; or\; segment\; [2;6][2;6],\; the\; intersection\; will\; become\; [2;4][2;4]\; (length\; 22)\; or\; [1;3][1;3](length\; 22),\; respectively.\; Removing\; any\; other\; segment\; will\; result\; in\; the\; intersection\; [2;3][2;3]\; (length\; 11).$

In the third example the intersection will become an empty set no matter the segment you remove.

In the fourth example you will get the intersection $[3;10][3;10]\; (length\; 77)\; if\; you\; remove\; the\; segment\; [1;5][1;5]\; or\; the\; intersection\; [1;5][1;5]\; (length\; 44)\; if\; you\; remove\; the\; segment\; [3;10][3;10].$

題意：給出n個區間，然後你可以刪除一個區間，問你剩下的區間的交集的最大長度是多少

思路：我們回到原始的求所有區間的交集，其實就是 距離左邊最近的右邊界-距離右邊最近的左邊界

所以這個問題我們可以排個序，然後我們枚舉要刪除的邊界，但是我們範圍是10^5 n^2算法肯定不行，但我們又需要遍歷，如果是nlogn的話肯定可以

這個時候我們可以考慮set,內有自動排序的功能，也是由紅黑樹組成優化了時間復雜度，但是說可能會記錄重復的，那我們就要使用 multiset，和set的區別就在於能記錄重復

然後我們枚舉每個刪除的區間即可

#include <bits/stdc++.h>
using namespace std;
int l[300005], r[300005];
multiset<int> a, b;
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++){
        scanf("%d%d", &l[i], &r[i]);
        a.insert(l[i]);
        b.insert(r[i]);
    }
    int ans = 0;
    for(int i=1;i<=n;i++){
        a.erase(a.find(l[i])); 
        b.erase(b.find(r[i]));
        ans = max(ans, *b.begin()-*a.rbegin());//取最大
        a.insert(l[i]);
        b.insert(r[i]);
    }
    printf("%d\n", ans);
}

還有種 優先隊列的寫法更加快

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

typedef long long ll;

int main() {
    int n,l[300010],r[300010];
    priority_queue<int> ml, mr;
    scanf("%d", &n);
    for (int i=0;i<n;++i) {
        scanf("%d %d", l+i, r+i);
        ml.push(l[i]);
        mr.push(-r[i]);
    }
    int ans = 0;
    bool bl, br;
    for (int i=0;i<n;++i) {
        bl = br = 0;
        if (ml.top()==l[i]) {
            bl = 1;
            ml.pop();
        }
        if (mr.top()==-r[i]) {
            br = 1;
            mr.pop();
        }
        ans = max(ans, -mr.top()-ml.top());
        if (bl) ml.push(l[i]);
        if (br) mr.push(-r[i]);
    }
    printf("%d\n", ans);
    return 0;
}

Codeforces Round #506 (Div. 3) C. Maximal Intersection