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题意：给定 n 个商店，然后有 m个限制，去 c 之前必须先去d，问你从0到n+1，最短路程是多少。

析：我们我们要到c，必须要先到d，那么举个例子，2 5， 3 7，如果我们先到5再到2，再到7再到3，那么3-5这个区间我们走了4次，如果我们先到7再到2，

那么就只走了3次，这很明显是最优的，所以我们把能合并的区间都合并起来，然后再一块计算。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#define debug puts("+++++")

//#include <tr1/unordered_map>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

//using namespace std :: tr1;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const LL LNF = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 2e5 + 5;

const LL mod = 1e9 + 7;

const int N = 1e6 + 5;

const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};

const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};

const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }

inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }

inline int lcm(int a, int b){  return a * b / gcd(a, b); }

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

struct Node{

    int l, r;

    bool operator < (const Node &p) const{

        return l < p.l || (l == p.l && r < p.r);

    }

};

Node a[505];

int main(){

    while(scanf("%d %d", &n, &m) == 2){

        for(int i = 0; i < m; ++i)  scanf("%d %d", &a[i].l, &a[i].r);

        if(!m){  printf("%d\n", n+1);  continue; }

        sort(a, a+m);

        int l = a[0].l, r = a[0].r;

        int ans = l;

        for(int i = 1; i < m; ++i){

            if(a[i].l <= r)  r = Max(r, a[i].r);

            else{ ans += 3 * (r-l) + a[i].l - r; l = a[i].l, r = a[i].r; }

        }

        ans += 3 * (r-l) + n - r + 1;

        printf("%d\n", ans);

    }

    return 0;

}

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