HDU - 5405 Sometimes Naive 树链剖分 （看题解）

#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 = 1e5 + 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;}

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, q, now;
LL w[N], sum[N], val[N];
vector<int> G[N];

int depth[N], top[N], son[N], sz[N], pa[N];
int in[N], ot[N], rk[N], idx;

struct Bit {
    int a[N];
    void init() {
        for(int i = 1; i <= n; i++) {
            a[i] = 0;
        }
    }
    void modify(int x, int v) {
        for(int i = x; i <= n; i += i & -i) {
            add(a[i], v);
        }
    }
    int sum(int x) {
        int ans = 0;
        for(int i = x; i; i -= i & -i) {
            add(ans, a[i]);
        }
        return ans;
    }
    int query(int L, int R) {
        if(L > R) return 0;
        return (sum(R) - sum(L - 1) + mod) % mod;
    }
} bit;

struct SegmentTree {
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
    int a[N << 2];
    void build(int l, int r, int rt) {
        if(l == r) {
            a[rt] = val[rk[l]];
            return;
        }
        int mid = l + r >> 1;
        build(lson); build(rson);
        a[rt] = a[rt << 1] + a[rt << 1 | 1];
        if(a[rt] >= mod) a[rt] -= mod;
    }
    void update(int p, int val, int l, int r, int rt) {
        if(l == r) {
            add(a[rt], val);
            return;
        }
        int mid = l + r >> 1;
        if(p <= mid) update(p, val, lson);
        else update(p, val, rson);
        a[rt] = a[rt << 1] + a[rt << 1 | 1];
        if(a[rt] >= mod) a[rt] -= mod;
    }
    int 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 a[rt];
        int mid = l + r >> 1;
        return (query(L, R, lson) + query(L, R, rson)) % mod;
    }
} Tree;


void dfs(int u, int fa) {
    pa[u] = fa;
    sz[u] = 1;
    sum[u] = w[u];
    depth[u] = depth[fa] + 1;
    for(auto &v : G[u]) {
        if(v == fa) continue;
        dfs(v, u);
        sz[u] += sz[v];
        add(sum[u], sum[v]);
        if(sz[son[u]] < sz[v]) {
            son[u] = v;
        }
    }
}

void dfs2(int u, int fa, int from) {
    in[u] = ++idx;
    rk[idx] = u;
    top[u] = from;
    if(son[u]) dfs2(son[u], u, from);
    for(auto &v : G[u]) {
        if(v == fa || v == son[u]) continue;
        dfs2(v, u, v);
        add(val[u], 1LL * sum[v] * sum[v] % mod);
    }
    ot[u] = idx;
}

inline int SUM(int u) {
    if(!u) return 0;
    return bit.query(in[u], ot[u]);
}

void update(int u, int val) {
    int fu = top[u];
    while(fu) {
        if(pa[fu]) {
            int now = SUM(fu);
            int pre = (now - val + mod) % mod;
            Tree.update(in[pa[fu]], mod - 1LL * pre * pre % mod, 1, n, 1);
            add(sum[fu], val);
            Tree.update(in[pa[fu]], 1LL * now * now % mod, 1, n, 1);
        }
        u = pa[fu]; fu = top[u];
    }
}


int query(int u, int v) {
    int ans = 0;
    int fu = top[u], fv = top[v];
    while(fu != fv) {
        if(depth[fu] < depth[fv]) {
            add(ans, Tree.query(in[fu], in[u], 1, n, 1));
            add(ans, 1LL * SUM(son[u]) * SUM(son[u]) % mod);
            if(pa[fu]) {
                sub(ans, 1LL * SUM(fu) * SUM(fu) % mod);
            }
            u = pa[fu]; fu = top[u];
        }
        else {
            add(ans, Tree.query(in[fv], in[v], 1, n, 1));
            add(ans, 1LL * SUM(son[v]) * SUM(son[v]) % mod);
            if(pa[fv]) {
                sub(ans, 1LL * SUM(fv) * SUM(fv) % mod);
            }
            v = pa[fv]; fv = top[v];
        }
    }
    if(depth[u] <= depth[v]) {
        add(ans, Tree.query(in[u], in[v], 1, n, 1));
        add(ans, 1LL * SUM(son[v]) * 1LL * SUM(son[v]) % mod);
        int tmp = (now - bit.query(in[u], ot[u]) + mod) % mod;
        add(ans, 1LL * tmp * tmp % mod);
    }
    else {
        add(ans, Tree.query(in[v], in[u], 1, n, 1));
        add(ans, 1LL * SUM(son[u]) * 1LL * SUM(son[u]) % mod);
        int tmp = (now - bit.query(in[v], ot[v]) + mod) % mod;
        add(ans, 1LL * tmp * tmp % mod);
    }
    return ans;
}

void init() {
    idx = 0;
    bit.init();
    for(int i = 1; i <= n; i++) {
        G[i].clear();
        val[i] = 0;
        sum[i] = 0;
        son[i] = 0;
        sz[i] = 0;
    }
}

int main() {
    while(scanf("%d%d", &n, &q) != EOF) {
        init();
        for(int i = 1; i <= n; i++) {
            scanf("%lld", &w[i]);
        }
        for(int i = 1; i < n; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }

        dfs(1, 0);
        dfs2(1, 0, 1);

        for(int i = 1; i <= n; i++) {
            bit.modify(in[i], w[i]);
        }


        now = sum[1];

        Tree.build(1, n, 1);
        while(q--) {
            int op, u, v;
            scanf("%d%d%d", &op, &u, &v);
            if(op == 1) {
                int tmp = (v - w[u] + mod) % mod;
                sub(now, w[u]);
                bit.modify(in[u], mod - w[u]);
                w[u] = v;
                add(now, w[u]);
                bit.modify(in[u], w[u]);
                update(u, tmp);
            }
            else {
                printf("%lld\n", (1LL * now * now % mod - query(u, v) + mod) % mod);
            }
        }
    }
    return 0;
}

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