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  • Codeforces 916E Jamie and Tree 线段树

    Jamie and Tree

    我们按 1 号点为根建dfs序线段树。即使换了根, 我们也能通过分类讨论去更新求答案。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 5e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}
    
    int n, q, a[N];
    int idx, dfn[N], in[N], ot[N];
    int depth[N], pa[N][20];
    vector<int> G[N];
    
    struct SegmentTree {
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
        LL sum[N << 2], lazy[N << 2];
        inline void pull(int rt) {
            sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
        }
        inline void push(int rt, int l, int r) {
            if(lazy[rt]) {
                int mid = l + r >> 1;
                sum[rt << 1] += lazy[rt] * (mid - l + 1);
                sum[rt << 1 | 1] += lazy[rt] * (r - mid);
                lazy[rt << 1] += lazy[rt];
                lazy[rt << 1 | 1] += lazy[rt];
                lazy[rt] = 0;
            }
        }
        void update(int L, int R, LL val, int l, int r, int rt) {
            if(R < l || r < L || R < L) return;
            if(L <= l && r <= R) {
                sum[rt] += val * (r - l + 1);
                lazy[rt] += val;
                return;
            }
            push(rt, l, r);
            int mid = l + r >> 1;
            update(L, R, val, lson);
            update(L, R, val, rson);
            pull(rt);
        }
        LL query(int L, int R, int l, int r, int rt) {
            if(R < l || r < L || R < L) return 0;
            if(L <= l && r <= R) return sum[rt];
            push(rt, l, r);
            int mid = l + r >> 1;
            return query(L, R, lson) + query(L, R, rson);
        }
    } Tree;
    
    void dfs(int u, int fa) {
        in[u] = ++idx;
        depth[u] = depth[fa] + 1;
        pa[u][0] = fa;
        for(int i = 1; i < 20; i++)
            pa[u][i] = pa[pa[u][i - 1]][i - 1];
        for(auto& v : G[u])
            if(v != fa) dfs(v, u);
        ot[u] = idx;
    }
    int getLca(int u, int v) {
        if(depth[u] < depth[v]) swap(u, v);
        int dis = depth[u] - depth[v];
        for(int i = 19; i >= 0; i--)
            if(dis >> i & 1) u = pa[u][i];
        if(u == v) return u;
        for(int i = 19; i >= 0; i--)
            if(pa[u][i] != pa[v][i])
                u = pa[u][i], v = pa[v][i];
        return pa[u][0];
    }
    
    int go(int u, int dis) {
        for(int i = 19; i >= 0; i--)
            if(dis >> i & 1) u = pa[u][i];
        return u;
    }
    
    bool ok(int i, int j) {
        return in[i] <= in[j] && ot[j] <= ot[i];
    }
    
    int main() {
        scanf("%d%d", &n, &q);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(int i = 2; i <= n; i++) {
            int u, v; scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        dfs(1, 0);
        for(int i = 1; i <= n; i++)
            Tree.update(in[i], in[i], a[i], 1, n, 1);
        int Rt = 1;
        while(q--) {
            int op, u, v, w, x;
            scanf("%d", &op);
            if(op == 1) {
                scanf("%d", &v);
                Rt = v;
            } else if(op == 2) {
                scanf("%d%d%d", &u, &v, &x);
                if(ok(Rt, u) && ok(Rt, v)) {
                    int lca = getLca(u, v);
                    if(lca != Rt) Tree.update(in[lca], ot[lca], x, 1, n, 1);
                    else Tree.update(1, n, x, 1, n, 1);
                } else if(ok(Rt, u)) {
                    Tree.update(1, n, x, 1, n, 1);
                } else if(ok(Rt, v)) {
                    Tree.update(1, n, x, 1, n, 1);
                } else {
                    w = getLca(u, v);
                    int lca = getLca(Rt, w);
                    if(lca == w) {
                        int p1 = getLca(Rt, u);
                        int p2 = getLca(Rt, v);
                        if(depth[p1] > depth[p2])  w = p1;
                        else w = p2;
                        int who = go(Rt, depth[Rt] - depth[w] - 1);
                        Tree.update(1, in[who] - 1, x, 1, n, 1);
                        Tree.update(ot[who] + 1, n, x, 1, n, 1);
                    } else {
                        Tree.update(in[w], ot[w], x, 1, n, 1);
                    }
                }
            } else {
                scanf("%d", &w);
                LL ans = 0;
                if(w == Rt) ans = Tree.query(1, n, 1, n, 1);
                else {
                    if(ok(w, Rt)) {
                        int who = go(Rt, depth[Rt] - depth[w] - 1);
                        ans += Tree.query(1, in[who] - 1, 1, n, 1);
                        ans += Tree.query(ot[who] + 1, n, 1, n, 1);
                    } else {
                        ans = Tree.query(in[w], ot[w], 1, n, 1);
                    }
                }
                printf("%lld
    ", ans);
            }
        }
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10874634.html
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