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树链剖分板子

时间:2019-10-06 00:18:20      阅读:89      评论:0      收藏:0      [点我收藏+]

标签:i++   pread   getch   stream   ble   for   dig   continue   enable   

// luogu-judger-enable-o2
#include<iostream>
#define p1 p << 1
#define p2 p << 1 | 1
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int N = 100050;
ll sum[N<<3], add_t[N<<3];
int l_t[N<<3], r_t[N<<3];
ll read(void) {
    ll x = 0, f = 1;
    char c = getchar();
    while (!isdigit(c)) {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 3) + (x << 1) + c - '0';
        c = getchar();
    }
    return x * f;
}
int siz[N], top[N];
int f[N], dep[N], son[N];
int h[N], to[N<<1];
int ne[N<<1], tot;
ll w[N];
inline void add_e(int x,int y) {
    ne[++tot] = h[x];
    h[x] = tot;
    to[tot] = y;
}

void dfs1(int x,int fa) {
    dep[x] = dep[fa] + 1;
    siz[x] = 1;
    f[x] = fa;
    for (int i = h[x]; i; i = ne[i]) {
        int y = to[i];
        if (y == fa) continue;
        dfs1(y, x);
        siz[x] += siz[y];
        if (siz[y] > siz[son[x]]) son[x] = y;
    }
}

ll wt[N], cnt;
int id[N], Top[N];

void dfs2(int x,int topf) {
    id[x] = ++cnt;
    wt[cnt] = w[x];
    Top[x] = topf;
    if (!son[x]) return;
    dfs2(son[x], topf);
    for (int i = h[x]; i; i = ne[i]) {
        int y = to[i];
        if (y == f[x] || y == son[x]) continue;
        dfs2(y, y);
    }
}

ll n, m, r, P;
void buil(int l,int r,int p) { 
    l_t[p] = l, r_t[p] = r;
    if (l == r) {
        sum[p] = wt[l] % P;
        return;
    }
    int mid = (l + r) >> 1;
    buil(l, mid, p1);
    buil(mid + 1, r, p2);
    sum[p] = sum[p1] + sum[p2];
}

void spread(int p) {
    if (add_t[p]) {
        sum[p1] += (ll)(r_t[p1] - l_t[p1] + 1) * add_t[p];
        sum[p2] += (ll)(r_t[p2] - l_t[p2] + 1) * add_t[p];
        add_t[p1] += add_t[p];
        add_t[p2] += add_t[p];
        sum[p1] %= P;
        sum[p2] %= P;
        add_t[p1] %= P;
        add_t[p2] %= P;
        add_t[p]= 0;
    }
}

void add(int l,int r,ll d,int p) {
    if (l_t[p] >= l && r_t[p] <= r) {
        sum[p] += (r_t[p] - l_t[p] + 1) * d;
        sum[p] %= P;
        add_t[p] += d;
        return;
    }
    spread(p);
    if (l <= r_t[p1]) add(l, r, d, p1);
    if (r >= l_t[p2]) add(l, r, d, p2);
    sum[p] = sum[p1] + sum[p2];
    sum[p] %= P;
}

ll ask(int l,int r,int p) {
    if (l_t[p] >= l && r_t[p] <= r)
        return sum[p];
    spread(p);
    ll ans = 0;
    if (r_t[p1] >= l) ans += ask(l, r, p1);
    if (l_t[p2] <= r) ans += ask(l, r, p2);
    return ans % P;
}

void update(int x,int y,ll z) {
    while (Top[x] != Top[y]) {
        if (dep[Top[x]] < dep[Top[y]]) swap(x, y);
        add(id[Top[x]], id[x], z, 1);
        x = f[Top[x]];
    }
    if (dep[x] < dep[y]) swap(x, y);
    add(id[y], id[x], z, 1);
}

ll query(int x,int y) {
    ll ans = 0;
    while (Top[x] != Top[y]) {
        if (dep[Top[x]] < dep[Top[y]]) swap(x, y);
        ans += ask(id[Top[x]], id[x], 1);
        x = f[Top[x]];
    }
    if (dep[x] < dep[y]) swap(x, y);
    ans += ask(id[y], id[x], 1);
    return ans % P;
}

void add_w(int x,ll z) {
    add(id[x], id[x] + siz[x] - 1, z, 1);
}

ll query_w(int x) {
    return ask(id[x], id[x] + siz[x] - 1, 1) % P;
}
    
int main() {
    n = read(), m = read(), r = read(), P = read();
    for (int i = 1;i <= n; i++) w[i] = read();
    for (int i = 1;i < n; i++) {
        int x = read(), y = read();
        add_e(x, y); add_e(y, x);
    }
    dfs1(r, 0);
    dfs2(r, r);
    buil(1, n, 1);
    for (int i = 1;i <= m; i++) {
        int op = read();
        if (op == 1) {
            int x = read(), y = read();
            ll z = read() % P;
            update(x, y, z);
        }
        else if (op == 2) {
            int x = read(), y = read();
            printf ("%lld\n", query(x, y));
        }
        else if (op == 3) {
            int x = read();
            ll z = read();
            add_w(x, z % P);
        }
        else {
            int x = read();
            printf ("%lld\n", query_w(x));
        }
    }
    return 0;
}

树链剖分板子

标签:i++   pread   getch   stream   ble   for   dig   continue   enable   

原文地址:https://www.cnblogs.com/Hs-black/p/11626115.html

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