[BZOJ3669][NOI2014]魔法森林（LCT）

直接求解比较麻烦。考虑先按照$A_i$为关键字将每条边排序，然后按顺序动态加入每一条边。以下定一条路径的代价为一条路径经过的所有边的$B$的最大值。可以知道，加入第$i$条边之后的最小代价，就是$A_i$加上从$1$到$n$的所有路径中，代价最小的路径的代价。
看到动态加边，可以想到使用LCT来完成。由于要维护边权，所以要把边在LCT中理解成点。
但是现在的一个问题是出现环时应该怎样处理。假设现在要连边$(u,v)$，但$u$和$v$之间已经存在一条路径。这样就必须在$u$到$v$的路径中选一条边断开。由于操作后要求得最小代价，因此，应该在$u$到$v$的路径中，选择一条边权最大的边断开后再连边$(u,v)$。当然，如果$u$和$v$在连边之前不连通，就不必要做这个步骤了。
连边之后，即可询问$1$到$n$的路径代价，加上$A_i$并更新答案。
代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
inline int read() {
    int res = 0; bool bo = 0; char c;
    while (((c = getchar()) < '0' || c > '9') && c != '-');
    if (c == '-') bo = 1; else res = c - 48;
    while ((c = getchar()) >= '0' && c <= '9')
        res = (res << 3) + (res << 1) + (c - 48);
    return bo ? ~res + 1 : res;
}
const int N = 2e5 + 5;
int n, m, fa[N], lc[N], rc[N], rev[N], que[N], len, val[N], sm[N],
f[N];
struct cyx {int a, b, x, y;} ask[N];
int cx(int x) {
    if (f[x] != x) f[x] = cx(f[x]);
    return f[x];
}
void zm(int x, int y) {
    int ix = cx(x), iy = cx(y);
    if (ix != iy) f[iy] = ix;
}
bool comp(cyx a, cyx b) {return a.y < b.y;}
int which(int x) {return rc[fa[x]] == x;}
bool is_root(int x) {
    return !fa[x] || (lc[fa[x]] != x && rc[fa[x]] != x);
}
void upt(int x) {
    sm[x] = x;
    if (lc[x] && val[sm[lc[x]]] > val[sm[x]]) sm[x] = sm[lc[x]];
    if (rc[x] && val[sm[rc[x]]] > val[sm[x]]) sm[x] = sm[rc[x]];
}
void down(int x) {
    if (rev[x]) {
        swap(lc[x], rc[x]);
        if (lc[x]) rev[lc[x]] ^= 1;
        if (rc[x]) rev[rc[x]] ^= 1;
        rev[x] = 0;
    }
} 
void rotate(int x) {
    int y = fa[x], z = fa[y], b = lc[y] == x ? rc[x] : lc[x];
    if (z && !is_root(y)) (lc[z] == y ? lc[z] : rc[z]) = x;
    fa[x] = z; fa[y] = x; b ? fa[b] = y : 0;
    if (lc[y] == x) rc[x] = y, lc[y] = b;
    else lc[x] = y, rc[y] = b; upt(y); upt(x);
}
void splay(int x) {
    int i, y; que[len = 1] = x;
    for (y = x; !is_root(y); y = fa[y]) que[++len] = fa[y];
    for (i = len; i >= 1; i--) down(que[i]);
    while (!is_root(x)) {
        if (!is_root(fa[x])) {
            if (which(x) == which(fa[x])) rotate(fa[x]);
            else rotate(x);
        }
        rotate(x);
    }
    upt(x);
}
void Access(int x) {
    int y;
    for (y = 0; x; y = x, x = fa[x]) {
        splay(x); rc[x] = y;
        if (y) fa[y] = x; upt(x);
    }
}
int Find_Root(int x) {
    Access(x); splay(x);
    while (down(x), lc[x]) x = lc[x];
    splay(x); return x;
}
void Make_Root(int x) {
    Access(x); splay(x);
    rev[x] ^= 1;
}
void Link(int x, int y) {
    Make_Root(x); fa[x] = y;
}
void Cut(int x, int y) {
    Make_Root(x); Access(y); splay(y);
    lc[y] = 0; fa[x] = 0; upt(y);
}
int Select(int x, int y) {
    Make_Root(x); Access(y); splay(y);
    return sm[y];
}
int main() {
    int i, ans = 2e9; n = read(); m = read();
    for (i = 1; i <= m; i++)
        ask[i].a = read(), ask[i].b = read(),
        ask[i].x = read(), ask[i].y = read();
    sort(ask + 1, ask + m + 1, comp);
    for (i = 1; i <= n + m; i++) f[i] = sm[i] = i;
    for (i = n + 1; i <= n + m; i++) val[i] = ask[i - n].x;
    for (i = 1; i <= m; i++) {
        int u = ask[i].a, v = ask[i].b; bool flag = 1;
        if (cx(u) == cx(v)) {
            int w = Select(u, v);
            if (val[w] > ask[i].x)
                Cut(ask[w - n].a, w), Cut(w, ask[w - n].b);
            else flag = 0;
        }
        else zm(u, v); if (flag) Link(u, i + n), Link(i + n, v);
        if (cx(1) == cx(n))
            ans = min(ans, ask[i].y + val[Select(1, n)]);
    }
    if (ans < 2e9) printf("%d\n", ans);
    else printf("-1\n");
    return 0;
}