题目地址：http://codeforces.com/problemset/problem/1029/C
题目：
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].
思路：
暴力贪心法。
给你一些集合，你要从中删去1个，使剩下集合的交集最大。
于是一个一个删去就行了。
顺便说一下multiset中的end()和rbegin()是不一样的恩。。。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<set>
#include<cmath>
using namespace std;
int l[300001],r[300001];
multiset<int>m1,m2;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
memset(l,0,sizeof(l));
memset(r,0,sizeof(r));
m1.clear();
m2.clear();
for(int i=0;i<n;i++)
{
scanf("%d%d",&l[i],&r[i]);
m1.insert(l[i]);
m2.insert(r[i]);
}
int ans=0;
for(int i=0;i<n;i++)
{
m1.erase(m1.find(l[i]));
m2.erase(m2.find(r[i]));
//cout<<*m2.begin()<<" "<<*m1.end()<<endl;
ans=max(ans,*m2.begin()-*m1.rbegin());
m1.insert(l[i]);
m2.insert(r[i]);
}
cout<<ans<<endl;
}
return 0;
}