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;

}

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