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：给出n个区间，从中删除一个区间，求剩下的区间的交集长度最大是多少
分析：根据区间求交集的性质，同大取大，同小取小，故求最大的左边界L1和次大的左边界L2，求最小的右边界R1和次小的右边界R2。
1、如果L1和R1都是题中所给的一个区间的左右边界，那么就把它删除，剩下的R2-L2就是答案
2、如果L1和R1分别处于题中所给的两个区间的端点，那么可以删L1,ans1=R1-L2, 或者删L2，ans2=R2-L1,ans=max(ans1,ans2)
AC code:
#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
struct Node{
int x,y;
}node[300005];
int main()
{
int n;
scanf("%d",&n);
int l1=-1,l2=-1,r1=INF,r2=INF;
for(int i=1;i<=n;i++){
scanf("%d%d",&node[i].x,&node[i].y);
int x=node[i].x,y=node[i].y;
if(l1<x) swap(l1,x);
if(l2<x) swap(l2,x);
if(r1>y) swap(r1,y);
if(r2>y) swap(r2,y);
}
int ans=0;
int l,r;
for(int i=1;i<=n;i++)
{
if(node[i].x==l1) l=l2;
else l=l1;
if(node[i].y==r1) r=r2;
else r=r1;
ans=max(ans,max(0,r-l));
}
printf("%d\n",ans);
return 0;
}