Segment

Description

They are N segments in x-axis, each segment has a start point X, a end point Y, so it can express as [X, Y].

Now you should select two segments, they have the longest overlap length.

Segment [3, 10] and segment [7, 12] overlap in [7, 10], so the length is 3.

Input Description

The first line is case number T( T <= 10 ).
For each case, the first line is the segment number N( 1 <= N <= 100,000 ), then follow N line, each line contian Xi and Yi.( 0 <= Xi < Yi < 100,000 )

Output Description

The longest overlap length of any two segments.

Sample Input

151 52 42 83 77 9

Sample Output

4

哎。如此水的一道题。我当时既然做不出来。。。。先按边起点从小到大排序，终点也从小到大排序。注意每条边都要喝前面的终点最右的边进行比较。

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;inline int max(int a,int b){return a>b?a:b;}struct Edge{int from,to;}edge[100008];bool cmp(Edge a,Edge b){if(a.from<b.from)return 1;if(a.from>b.from)return 0;if(a.from==b.from) {if(a.to>b.to)return 0;else return 1;}}int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d%d",&edge[i].from,&edge[i].to);}sort(edge+1,edge+n+1,cmp);int s=edge[1].from,t=edge[1].to;int maxlen=0;for(int i=2;i<=n;i++){if(edge[i].from<t){if(edge[i].to<=t){maxlen=max(maxlen,edge[i].to-edge[i].from);}else {maxlen=max(maxlen,t-edge[i].from);s=edge[i].from;t=edge[i].to;}}else continue;}printf("%d\n",maxlen);}return 0;}