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Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

# Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) A. Math Problem 水题

## A. Math Problem

Your math teacher gave you the following problem:

There are n segments on the x-axis, [l1;r1],[l2;r2],…,[ln;rn]. The segment [l;r] includes the bounds, i.e. it is a set of such x that l≤x≤r. The length of the segment [l;r] is equal to r−l.

Two segments [a;b] and [c;d] have a common point (intersect) if there exists x that a≤x≤b and c≤x≤d. For example, [2;5] and [3;10] have a common point, but [5;6] and [1;4] don't have.

You should add one segment, which has at least one common point with each of the given segments and as short as possible (i.e. has minimal length). The required segment can degenerate to be a point (i.e a segment with length zero). The added segment may or may not be among the given n segments.

In other words, you need to find a segment [a;b], such that [a;b] and every [li;ri] have a common point for each i, and b−a is minimal.

## Input

The first line contains integer number t (1≤t≤100) — the number of test cases in the input. Then t test cases follow.

The first line of each test case contains one integer n (1≤n≤105) — the number of segments. The following n lines contain segment descriptions: the i-th of them contains two integers li,ri (1≤li≤ri≤109).

The sum of all values n over all the test cases in the input doesn't exceed 105.

## Output

For each test case, output one integer — the smallest possible length of the segment which has at least one common point with all given segments.

input
4
3
4 5
5 9
7 7
5
11 19
4 17
16 16
3 12
14 17
1
1 10
1
1 1
output
2
4
0
0

## Note

In the first test case of the example, we can choose the segment [5;7] as the answer. It is the shortest segment that has at least one common point with all given segments.

## 代码

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

vector<pair<int,int> >A;
void solve(){
int n;cin>>n;
A.clear();
for(int i=0;i<n;i++){
int x,y;
cin>>x>>y;
A.push_back(make_pair(x,y));
}
sort(A.begin(),A.end());
int ll = A[0].first,lr=A[0].second;
int rl = A[0].first,rr=A[0].second;
for(int i=0;i<n;i++){
lr = min(lr,A[i].second);
rl = max(rl,A[i].first);
}
// cout<<lr<<" "<<rl<<endl;
if(lr>=rl){
cout<<"0"<<endl;
}else{
cout<<rl-lr<<endl;
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
solve();
}
}
posted @ 2019-11-24 22:59  qscqesze  阅读(148)  评论(0编辑  收藏