Codeforces 838 B

B - Diverging Directions

思路：

用dfs序+线段树维护子树中距离（从1到u，再从u到1）的最小值

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 2e5 + 5;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int L[N], R[N], anc[N][20], deep[N], f[N], b[N], bb[N], head[N], cnt = 1, now = 1;
LL tree[N<<2], lazy[N<<2], a[N];
struct edge {
    int from, to, w, nxt;
}edge[N];
void add_edge(int u, int v, int w) {
    edge[cnt].from = u;
    edge[cnt].to = v;
    edge[cnt].w = w;
    edge[cnt].nxt = head[u];
    head[u] = cnt++;
}

void push_up(int rt) {
    tree[rt] = min(tree[rt<<1], tree[rt<<1|1]);
}
void push_down(int rt) {
    lazy[rt<<1] += lazy[rt];
    lazy[rt<<1|1] += lazy[rt];
    tree[rt<<1] += lazy[rt];
    tree[rt<<1|1] += lazy[rt];
    lazy[rt] = 0;
}
void build(int rt, int l, int r) {
    if(l == r) {
        tree[rt] = a[f[l]];
        return ;
    }
    int m = l+r >> 1;
    build(ls);
    build(rs);
    push_up(rt);
}
void update(int L, int R, int v, int rt, int l, int r) {
    if(L <= l && r <= R) {
        tree[rt] += v;
        lazy[rt] += v;
        return ;
    }
    if(lazy[rt]) push_down(rt);
    int m = l+r >> 1;
    if(L <= m) update(L, R, v, ls);
    if(R > m) update(L, R, v, rs);
    push_up(rt);
}
LL query(int L, int R, int rt, int l, int r) {
    if(L <= l && r <= R) return tree[rt];
    if(lazy[rt]) push_down(rt);
    int m = l+r >> 1;
    LL ans = INF;
    if(L <= m) ans = min(ans, query(L, R, ls));
    if(R > m) ans = min(ans, query(L, R, rs));
    push_up(rt);
    return ans;
}
void dfs(int u) {
    L[u] = now;
    f[now] = u;
    for (int i = head[u]; ~i; i = edge[i].nxt) {
        int v = edge[i].to;
        anc[v][0] = u;
        for (int j = 1; j < 20; j++) anc[v][j] = anc[anc[v][j-1]][j-1];
        deep[v] = deep[u] + 1;
        now++;
        a[v] = a[u] - b[u] + edge[i].w + b[v];
        dfs(v);
    }
    R[u] = now;
}
int lca(int u, int v) {
    if(deep[u] < deep[v]) swap(u, v);
    for (int i = 19; i >= 0; i--) if(deep[anc[u][i]] >= deep[v]) u = anc[u][i];
    if(u == v) return u;
    for (int i = 19; i >= 0; i--) if(anc[u][i] != anc[v][i]) u = anc[u][i], v = anc[v][i];
    return anc[u][0];
}
int main() {
    int n, q, ty, u, v, w;
    mem(head, -1);
    scanf("%d %d", &n, &q);
    for (int i = 1; i < n; i++) {
        scanf("%d %d %d", &u, &v, &w);
        add_edge(u, v, w);
    }

    for (int i = 1; i < n; i++) {
        scanf("%d %d %d", &u, &v, &w);
        b[u] = w;
        bb[i] = u;
    }

    for (int i = 0; i < 20; i++) anc[1][i] = 1;
    dfs(1);
    build(1, L[1], R[1]);
    while(q--) {
        scanf("%d %d %d", &ty, &u, &v);
        if(ty == 1) {
            if(u >= n) {
                int nod = bb[u-n+1];
                int add = v - b[nod];
                update(L[nod], L[nod], add, 1, L[1], R[1]);
                b[nod] = v;
            }
            else {
                int nod = edge[u].to;
                int add = v - edge[u].w;
                update(L[nod], R[nod], add, 1, L[1], R[1]);
                edge[u].w = v;
            }
        }
        else {
            int l = lca(u, v);
            LL ans = INF;
            if(l == u) {
                LL dis1 = query(L[u], L[u], 1, L[1], R[1]) - b[u];
                LL dis2 = query(L[v], L[v], 1, L[1], R[1]) - b[v];
                ans = dis2 - dis1;
            }
            else {
                LL dis1 = query(L[u], R[u], 1, L[1], R[1]) - (query(L[u], L[u], 1, L[1], R[1]) - b[u]);
                LL dis2 = query(L[v], L[v], 1, L[1], R[1]) - b[v];
                ans = dis1 + dis2;
            }
            printf("%lld
", ans);
        }
    }
    return 0;
}