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

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 $\mathrm{00 in\; case\; the\; intersection\; is\; an\; empty\; set.}$

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 $\mathrm{nn (2\le n\le 3\cdot 1052\le n\le 3\cdot 105)\; —\; the\; number\; of\; segments\; in\; the\; sequence.}$

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

题意：给出n个区间，然后你可以删除一个区间，问你剩下的区间的交集的最大长度是多少

思路：我们回到原始的求所有区间的交集，其实就是      距离左边最近的右边界-距离右边最近的左边界

所以这个问题我们可以排个序，然后我们枚举要删除的边界，但是我们范围是10^5     n^2算法肯定不行，但我们又需要遍历，如果是nlogn的话肯定可以

这个时候我们可以考虑set,内有自动排序的功能，也是由红黑树组成优化了时间复杂度，但是说可能会记录重复的，那我们就要使用 multiset，和set的区别就在于能记录重复

然后我们枚举每个删除的区间即可

#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
", ans);
}

 还有种 优先队列的写法更加快

#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
", ans);
    return 0;
}