LOJ P1967 :货车运输 最大生成树 + 树链剖分

传送门

分析

跟上一题基本一样的思路
（我从哪翻出来这道题的我也忘了。。。）

代码

#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
struct Edge{
    int u,v,w;
}edge[M];
int q[N],n,m,cnt,qu;
int h[N],ne[M],e[M],w[M],idx;
int id[N], son[N], sz[N], fa[N], dep[N], top[N], son_w[N];
int nw[N];
bool st[N];
struct Node{
    int l,r;
    int res;
}tr[N * 4];

void add(int x,int y,int z){
    ne[idx] = h[x],e[idx] = y,w[idx] = z,h[x] = idx++;
}

bool cmp(Edge A,Edge B){
    return A.w > B.w;
}

int find(int x){
    if(x != q[x]) q[x] = find(q[x]);
    return q[x];
}

void dfs(int u, int f, int d) {
    st[u] = true;
    sz[u] = 1, son[u] = 0, fa[u] = f, dep[u] = d;
    for (int i  = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j == f) continue;
        dfs(j, u, d + 1);
        sz[u] += sz[j];
        if (sz[j] > sz[son[u]]) {
            son[u] = j;
            son_w[u] = w[i];
        }

    }
}

void dfs_(int u, int t, int res) {
    id[u] = ++cnt, nw[cnt] = res, top[u] = t;
    if (!son[u]) return;
    dfs_(son[u], t, son_w[u]);
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j == fa[u]) continue;
        if (j == son[u]) continue;
        else dfs_(j, j, w[i]);
    }
}

void build(int u,int l,int r){
    tr[u].l = l,tr[u].r = r;
    if(l == r){
        tr[u].res = nw[l];
        return;
    }
    int mid = l + r >> 1;
    build(u << 1,l,mid),build(u << 1 | 1,mid + 1,r);
    tr[u].res = min(tr[u << 1].res,tr[u << 1 | 1].res);
}

int query(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) return tr[u].res;
    int mid = (tr[u].l + tr[u].r) >> 1;
    int res = 0x3f3f3f3f;
    if (l <= mid) res = query(u << 1, l, r);
    if (r > mid) res = min(res, query(u << 1 | 1, l, r));
    return res;
}

int query_path(int l, int r) {
    int res = 0x3f3f3f3f;
    while (top[l] != top[r]) {
        if (dep[top[l]] < dep[top[r]]) swap(l, r);
        res = min(query(1, id[top[l]], id[l]), res);
        l = fa[top[l]];
    }
    if (dep[l] > dep[r]) swap(l, r);
    res = min(res, query(1, id[son[l]], id[r]));
    return res;
}

int main() {
    read(n),read(m);
    for(int i = 1;i <= n;i++) h[i] = -1,q[i] = i,nw[i] = INF;
    for(int i = 1;i <= m;i++) read(edge[i].u),read(edge[i].v),read(edge[i].w);
    sort(edge + 1,edge + 1 + m,cmp);  
    for(int i = 1;i <= m;i++){
        int x = find(edge[i].u),y = find(edge[i].v);
        if(x != y){
            q[x] = y;
            add(edge[i].u,edge[i].v,edge[i].w);
            add(edge[i].v,edge[i].u,edge[i].w);
        }
    }
    for(int i = 1;i <= n;i++){
        if(!st[i]){
            dfs(i, -1, 1);
            dfs_(i, 1, 0);
        }
    }
    build(1, 1, n);  
    read(qu);
    while(qu--){
        int l,r;
        read(l),read(r);
        if(find(l) != find(r)) puts("-1");
        else di(query_path(l,r));
    }
    return 0;
}