LCT

和魔法森林类似，先排序一个关键字，用LCT动态维护另一个关键字。

这里把r从大到小排序，然后动态加边。

因为r的顺序是确定的，每次加入新边以后，r肯定是看加入的这条边的，所以我们考虑l就好了。

对于l，我们要求能够通过的数量，肯定是要保留合法的大范围，因为小范围包含在了大范围里。

之后就和魔法森林一样维护1～n的链。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define __fastIn ios::sync_with_stdio(false), cin.tie(0)#define pb push_backusing namespace std;typedef long long LL;inline int lowbit(int x){ return x & (-x); }inline int read(){    int ret = 0, w = 0; char ch = 0;    while(!isdigit(ch)){        w |= ch == '-', ch = getchar();    }    while(isdigit(ch)){        ret = (ret << 3) + (ret << 1) + (ch ^ 48);        ch = getchar();    }    return w ? -ret : ret;}template 
     
      inline A __lcm(A a, A b){ return a / __gcd(a, b) * b; }template 
      
       inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 200005;int n, m, tot, ch[N][2], fa[N], rev[N], mx[N], id[N], st[N], w[N];struct Edge{    int u, v, l, r;    bool operator < (const Edge &rhs) const {        return r > rhs.r;    }}e[N];int build(int val){    ++ tot;    w[tot] = mx[tot] = val, id[tot] = tot;    ch[tot][0] = ch[tot][1] = rev[tot] = fa[tot] = 0;    return tot;}bool isRoot(int x){    return ch[fa[x]][0] != x && ch[fa[x]][1] != x;}void push_up(int rt){    int l = ch[rt][0], r = ch[rt][1];    id[rt] = rt, mx[rt] = w[rt];    if(mx[l] > mx[rt]) mx[rt] = mx[l], id[rt] = id[l];    if(mx[r] > mx[rt]) mx[rt] = mx[r], id[rt] = id[r];}void reverse(int x){    rev[x] ^= 1;    swap(ch[x][0], ch[x][1]);}void push_down(int x){    if(rev[x]){        int l = ch[x][0], r = ch[x][1];        if(l) reverse(l);        if(r) reverse(r);        rev[x] ^= 1;    }}void rotate(int x){    int y = fa[x], z = fa[y], p = (ch[y][1] == x) ^ 1;    ch[y][p ^ 1] = ch[x][p], fa[ch[x][p]] = y;    if(!isRoot(y)) ch[z][ch[z][1] == y] = x;    fa[x] = z, fa[y] = x, ch[x][p] = y;    push_up(y), push_up(x);}void splay(int x){    int pos = 0; st[++ pos] = x;    for(int i = x; !isRoot(i); i = fa[i]) st[++ pos] = fa[i];    while(pos) push_down(st[pos --]);    while(!isRoot(x)){        int y = fa[x], z = fa[y];        if(!isRoot(y)){            (ch[y][0] == x) ^ (ch[z][0] == y) ? rotate(x) : rotate(y);        }        rotate(x);    }    push_up(x);}void access(int x){    for(int p = 0; x; p = x, x = fa[x])        splay(x), ch[x][1] = p, push_up(x);}void makeRoot(int x){    access(x), splay(x), reverse(x);}void link(int x, int y){    makeRoot(x), fa[x] = y;}int findRoot(int x){    access(x), splay(x);    while(ch[x][0]) push_down(x), x = ch[x][0];    splay(x);    return x;}void split(int x, int y){    makeRoot(x), access(y), splay(y);}bool isConnect(int x, int y){    makeRoot(x);    return findRoot(y) == x;}vector
       
        
         > ans;int main(){    n = read(), m = read();    for(int i = 1; i <= m; i ++){        e[i].u = read(), e[i].v = read(), e[i].l = read(), e[i].r = read();    }    for(int i = 1; i <= n; i ++) build(0);    sort(e + 1, e + m + 1);    for(int i = 1; i <= m; i ++){        int u = e[i].u, v = e[i].v, t = build(e[i].l);        if(!isConnect(u, v)){            link(u, t), link(t, v);        }        else{            split(u, v);            if(mx[v] >= e[i].l){                int p = id[v];                splay(p);                fa[ch[p][0]] = 0, fa[ch[p][1]] = 0;                ch[p][0] = ch[p][1] = 0;                link(u, t), link(t, v);            }        }        if(isConnect(1, n)){            split(1, n);            if(mx[n] <= e[i].r) ans.pb(make_pair(mx[n], e[i].r));        }    }    sort(ans.begin(), ans.end());    int res = 0;    for(int i = 0; i < ans.size(); i ++){        int j = i, cur = ans[i].second;        while(j + 1 < ans.size() && ans[j + 1].first <= cur) cur = max(cur, ans[++ j].second);        res += cur - ans[i].first + 1;        i = j;    }    printf("%d\n", res);    return 0;}