LOJ 147 DFS序+ST表求LCA+树上差分

题目链接

树状数组支持区间加减的修改(似乎不支持乘除等复杂的修改)

①ST表忘了初始化和i、j写反

②我们维护的dfs序(也可维护欧拉序)是前序遍历，根节点总在前面，求ST表一定用的是欧拉序

③w数组和树状数组，add参数一定要ll！不然莫名WA

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 2000006;
struct {
    int to, next, w;
} e[maxn << 2];

int head[maxn << 1], edgeNum;
void add(int u, int v) {
    e[++edgeNum].to = v;
    e[edgeNum].next = head[u];
    head[u] = edgeNum;
}

int in[maxn], dfn[maxn], n, cnt, cnt2, m, R, deep[maxn], fa1[maxn];
ll w[maxn];
struct aaa {
    ll S[maxn];
    void add(int x, ll c) {
        while (x <= n) S[x] += c, x += x & -x;
    }
    ll Sum(int x) {
        ll res = 0;
        while (x > 0) res += S[x], x -= x & -x;
        return res;
    }
} st1, st2;
// ll S[3][maxn];
int siz[maxn];
void init() {
    // memset(head,-1,4*n+4);
    cnt = edgeNum = 0;
}

// ST求LCA
int minl[25][maxn], lg[maxn];
int tmp;
inline void S_table() {
    for (int i = 1; i <= cnt; ++i) lg[i] = (1 << (lg[i - 1] + 1)) <= i ? lg[i - 1] + 1 : lg[i - 1];
    for (int j = 1; (1 << j) <= cnt; ++j)
        for (register int i = 1; i + (1 << j) - 1 <= cnt; ++i) {
            minl[j][i] = deep[minl[j - 1][i]] < deep[minl[j - 1][i + (1 << (j - 1))]]
                             ? minl[j - 1][i]
                             : minl[j - 1][i + (1 << (j - 1))];
           
        }
}

inline int lca(int l, int r) {
    if (l > r)
        swap(l, r);
    int k = lg[r - l + 1];
    // int k = log2((double)(r-l+1));
    int mid = r - (1 << k) + 1;
    // return min(minl[l][k],minl[mid][k]);
    return deep[minl[k][l]] < deep[minl[k][r - (1 << k) + 1]] ? minl[k][l]
                                                                      : minl[k][r - (1 << k) + 1];
}



ll query(int x) {
    if (x == 0)
        return 0;

    return st1.Sum(dfn[x]) + (deep[x] + 1) * st2.Sum(dfn[x]) + w[x]; 
}
void dfs(int u) {
    in[u] = ++cnt;
    w[u] += w[fa1[u]];
    deep[u] = deep[fa1[u]] + 1;
    minl[0][cnt] = u;
    dfn[u] = ++cnt2;
    siz[u] = 1;
    for (int i = head[u]; i; i = e[i].next) {
        int v = e[i].to;
        if (v == fa1[u])
            continue;
        fa1[v] = u;
        dfs(v);
        // out[u] = ++cnt;
        minl[0][++cnt] = u;
        siz[u] += siz[v];
    }
}
int main() {
    scanf("%d%d%d", &n, &m, &R);
    for (int i = 1; i <= n; ++i) scanf("%lld", &w[i]);
    for (int i = 0, u, v; i < n - 1; ++i) scanf("%d%d", &u, &v), add(u, v), add(v, u);

    dfs(R);  //等于自己dfs会加多
    S_table();

    for (int i = 0, p, u, v; i < m; ++i) {
        scanf("%d%d%d", &p, &u, &v);  // 0单点更值
        if (p == 1)
            st1.add(dfn[u], v), st1.add(dfn[u] + siz[u], -v);
        else if (p == 2) {  // 1 值*dep
            st1.add(dfn[u], -1ll * v * (deep[u])), st1.add(dfn[u] + siz[u], 1ll * v * (deep[u]));
            st2.add(dfn[u], v), st2.add(dfn[u] + siz[u], -v);  // 2 个数
        } else {
            int L = lca(in[u], in[v]);
            //区间查询，树上差分
            printf("%lld\n", query(u) + query(v) - query(L) - query(fa1[L]));
        }
        // printf("%lld\n",query(u));//单点查询
    }
}