数据太水了，其实这个解法是错的

因为题目中要求sorted顺次连接而成，所以不能直接sort+贪心

正解：
#include<iostream>
#include<stack>
#include<string.h>
#include<cmath>
#include<iomanip>
#include<algorithm>
#include<climits>
#include<cstdio>
#include<vector>
#include<sstream>
#include<ctype.h>
#include<set>
#include<map>
#include<ctime>
#include<stdlib.h>
#include<queue>
#include<bitset>
#define ll long long
#define inf INT_MAX
using namespace std;
template<typename TTT>inline void mr(TTT& theNumberToRead)
{
	theNumberToRead = 0;
	bool prn = false;
	char c = getchar();
	while (!isdigit(c))
	{
		if (c == '-')prn = true;
		c = getchar();
	}
	while (isdigit(c))
		theNumberToRead = 10 * theNumberToRead + (c ^ 48), c = getchar();
	if (prn)
		theNumberToRead = -theNumberToRead;
}
template<typename TTT>inline TTT mrr()
{
	TTT theNumberToRead = 0;
	bool prn = false;
	char c = getchar();
	while (!isdigit(c))
	{
		if (c == '-')prn = true;
		c = getchar();
	}
	while (isdigit(c))
		theNumberToRead = 10 * theNumberToRead + (c ^ 48), c = getchar();
	return prn ? -theNumberToRead : theNumberToRead;
}
template<typename T>void my_write(T x)
{
	if (x)
		my_write(x / 10),
		putchar(x % 10 ^ 48);
}
template<typename T>void mw(T x, char mid)
{
	if (x)
	{
		if (x < 0)
			putchar('-'),
			x = -x;
		my_write(x);
	}
	else putchar(48);
	if (mid)
		putchar(mid);
}
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
// ******************************************华丽的分割线******************************************
const int maxm = 500000, maxn = 50000;
int s[maxm], t[maxm], dp[maxn + 1], n2, st[1 << 17];
inline int ls(int x)
{
	return x << 1;
}
inline int rs(int x)
{
	return x << 1 | 1;
}
inline int fa(int x)
{
	return x >> 1;
}
inline void up(int x)
{
	st[x] = min(st[ls(x)], st[rs(x)]);
}
void st_init(int n)
{
	for (n2 = 1; n2 <= n; n2 <<= 1);
	fill(st, st + (1 << 17), 500000);
	for (int i = n2 + 1; i; i >>= 1)
		st[i] = 0;
}
void update(int i, int x)
{
	i += n2, st[i] = x;
	for (i >>= 1; i; i >>= 1)
		up(i);
}
int query(int l, int r)
{
	int ans = INT_MAX;
	for (l += n2 - 1, r += n2 + 1; l ^ r ^ 1; l >>= 1, r >>= 1)
	{
		if (~l & 1 && ans > st[l ^ 1])
			ans = st[l ^ 1];
		if (r & 1 && ans > st[r ^ 1])
			ans = st[r ^ 1];
	}
	return ans;
}
int main()
{
	int n, m;
	mr(n), mr(m);
	for (int i = 0; i < m; ++i)
		mr(s[i]), mr(t[i]);
	fill(dp, dp + n + 1, 500000);
	dp[1] = 0;
	st_init(n);
	for (int i = 0, v; i < m; ++i)
	{
		v = min(dp[t[i]], query(s[i], t[i]) + 1);
		dp[t[i]] = v;
		update(t[i], v);
	}
	mw(dp[n], '\n');
	return 0;
}