题意：有n个长度为1的线段，确定它们的起点，使得第i个线段在【ri，di】之间，输出空隙数目的最小值。

思路：贪心。首先对区间进行排序，然后扫描一遍，当上一个线段最右边加1《=当前最左的点的时，那么ans++；

cdode：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=100005;
const int M=2005;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define fr(i,s,n) for (int i=s;i<=n;i++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);pii v[N];
int main()
{int T,n;scanf("%d",&T);while(T--){scanf("%d",&n);fr(i,0,n-1) scanf("%d %d",&v[i].first,&v[i].second);sort(v,v+n);int l=-INF,r=INF,ans=0;fr(i,0,n-1){l=max(l+1,v[i].first);r=min(r+1,v[i].second);if (l>=r){++ans;l=v[i].first;r=v[i].second;}}printf("%d\n",ans);}
}