传送门
题面：
C. Maximal Intersection
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given 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 nn (2≤n≤3⋅1052≤n≤3⋅105) — the number of segments in the sequence.
Each of the next nn lines contains two integers lili and riri (0≤li≤ri≤1090≤li≤ri≤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：
给你一个n个区间[l,r]，问当你删掉某1个区间后，剩下的n-1个区间所能够相交的长度的最大值。
题目分析：
首先，我们首先考虑将所有的左括号和所有的右括号从小到大排序，那么不难发现，倘若我们需要求多个区间的相交的长度，那么我们只能选取左括号中序号最大的和右括号中最小的，他们相减即为最大长度（若为负数则为0）。
而题目中要求我们删掉某个区间，我们于是我们就可以考虑枚举被删的区间。我们可以用multiset分别存储左右括号的坐标，在枚举的过程中先删除第i个区间，统计答案之后再加上第i个区间。不断遍历更新最大值即可。
代码：
#include <bits/stdc++.h>
#define maxn 300005
using namespace std;
multiset<int>l,r;
struct Seg{
int l,r;
}q[maxn];
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++){
scanf("%d%d",&q[i].l,&q[i].r);
l.insert(q[i].l);//用multiset存左右括号
r.insert(q[i].r);
}
int res=0;
for(int i=1;i<=n;i++){
l.erase(l.find(q[i].l));//删除第i个区间
r.erase(r.find(q[i].r));
res=max(res,*r.begin()-*--l.end());//答案是右区间的最小值和左区间的最大值
l.insert(q[i].l);
r.insert(q[i].r);
}
cout<<res<<endl;
}