BZOJ 3669 魔法森林

LCT维护生成树

先按照a的权值把边排序，离线维护b的最小生成树。

将a排序后，依次动态加边，我们只需要关注b的值。要保证1-n花费最少，两点间的b值肯定是越小越好，所以我们可以考虑以b为关键字维护最小生成树。

对于新加的边b，如果1-n已经联通，需要更新答案

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
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 = 400005;
int n, m, tot, ans, mx[N], fa[N], w[N], ch[N][2], rev[N], id[N], st[N];
struct Edge {
    int v, u, a, b;
    bool operator < (const Edge &rhs) const {
        return a < rhs.a;
    }
} e[N];

int newNode(int v){
    ++tot;
    w[tot] = mx[tot] = v, id[tot] = tot;
    fa[tot] = ch[tot][0] = ch[tot][1] = 0;
    return tot;
}

bool isRoot(int x){
    return ch[fa[x]][0] != x && ch[fa[x]][1] != x;
}

void reverse(int x){
    rev[x] ^= 1, swap(ch[x][0], ch[x][1]);
}

void push_up(int x){
    int l = ch[x][0], r = ch[x][1];
    mx[x] = w[x], id[x] = x;
    if(mx[l] > mx[x]) mx[x] = mx[l], id[x] = id[l];
    if(mx[r] > mx[x]) mx[x] = mx[r], id[x] = id[r];
}

void push_down(int x){
    if(rev[x]){
        rev[x] ^= 1;
        if(ch[x][0]) reverse(ch[x][0]);
        if(ch[x][1]) reverse(ch[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;
}

int main(){

    ans = INF;
    n = read(), m = read();
    for(int i = 1; i <= n; i ++) newNode(0);
    for(int i = 0; i < m; i ++){
        e[i].u = read(), e[i].v = read(), e[i].a = read(), e[i].b = read();
    }
    sort(e, e + m);
    for(int i = 0; i < m; i ++){
        int u = e[i].u, v = e[i].v, t = newNode(e[i].b);
        if(!isConnect(u, v)) link(u, t), link(t, v);
        else{
            split(u, v);
            if(e[i].b > mx[v]) continue;
            int tmp = id[v]; splay(tmp);
            fa[ch[tmp][0]] = fa[ch[tmp][1]] = 0;
            ch[tmp][0] = ch[tmp][1] = 0;
            link(u, t), link(t, v);
        }
        if(isConnect(1, n)){
            split(1, n);
            ans = min(ans, mx[n] + e[i].a);
        }
    }
    printf(ans == INF ? "-1\n" : "%d\n", ans);
    return 0;
}