洛谷p3384【模板】樹鏈剖分題解

洛谷p3384 【模板】樹鏈剖分錯誤記錄

首先感謝\(lfd\)在課上調了出來\(Orz\)

\(1\).之後少寫全局變量

\(2\).線段樹遞歸的時候最好把左右區間一塊兒傳

\(3\).寫\(dfs\)的時候不要寫錯名字

\(4\).使用線段樹的操做的時候纔要用到\(dfs\)序

\(5\).須要開一個數組來記錄在\(dfs\)序下的節點是什麼也方便線段樹的賦值

\(6\).注意\(down\)函數內怎樣更新

\(7\).在查詢的時候並不須要向上更新

因爲\(yxj\)看了\(lfd\)敲的樹鏈剖分感受壓完行以後很是的好看，由此\(yxj\)也踏上了瘋狂壓行的不歸路

Code：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define lson k << 1
#define rson k << 1 | 1
using namespace std;
const int N = 1e5+7;
int n, m, R, p, dfn[N], top[N], son[N], dep[N], fa[N], siz[N], tot, head[N << 1], cnt, num, x, y, z, w[N], l, r, ans, pre[N];
struct node {int l, r, f, w;}tr[N << 2];
struct Node {int nxt, to;}e[N << 1];
int read() {
    int s = 0, w = 1;
    char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') w = -1; ch = getchar();}
    while(isdigit(ch)) {s = s * 10 + ch - '0'; ch = getchar();}
    return s * w;
}
void build(int k, int l, int r) {
    tr[k].l = l, tr[k].r = r;
    if(l == r) {tr[k].w = w[pre[l]]; return;}
    int mid = (l + r) >> 1;
    build(lson, l, mid);
    build(rson, mid + 1, r);
    tr[k].w = (tr[lson].w + tr[rson].w) % p;
}
void add(int x, int y) {
    e[++cnt].nxt = head[x];
    e[cnt].to = y;
    head[x] = cnt;
}
void dfs(int x) {
    siz[x] = 1;
    dep[x] = dep[fa[x]] + 1;
    for(int i = head[x]; i; i = e[i].nxt) {
        if(e[i].to == fa[x]) continue;
        fa[e[i].to] = x, dfs(e[i].to), siz[x] += siz[e[i].to];
        if(siz[e[i].to] > siz[son[x]]) son[x] = e[i].to;
    }
} 
void dfs1(int x) {
    dfn[x] = ++tot; pre[tot] = x; 
    if(!top[x]) top[x] = x;
    if(son[x]) top[son[x]] = top[x], dfs1(son[x]);
    for(int i = head[x]; i; i = e[i].nxt) 
        if(e[i].to != fa[x] && e[i].to != son[x]) dfs1(e[i].to);
}
void down(int k) {
    tr[lson].f += tr[k].f; tr[rson].f += tr[k].f;
    tr[lson].w += (tr[lson].r - tr[lson].l + 1) * tr[k].f; 
    tr[rson].w += (tr[rson].r - tr[rson].l + 1) * tr[k].f;
    tr[k].f = 0;
}
void change_query(int k) {
    if(tr[k].l >= l && tr[k].r <= r) {tr[k].w += (tr[k].r - tr[k].l + 1) * z;tr[k].f += z; return;}
    if(tr[k].f) down(k);
    int mid = (tr[k].l + tr[k].r) >> 1;
    if(l <= mid) change_query(lson);
    if(r > mid) change_query(rson);
    tr[k].w = (tr[lson].w + tr[rson].w) % p;
}
void work(int x, int y) {
    z %= p;
    while(top[x] != top[y]) {
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        l = dfn[top[x]], r = dfn[x], change_query(1);
        x = fa[top[x]];
    }
    if(dep[x] > dep[y]) swap(x, y);
    l = dfn[x], r = dfn[y]; change_query(1);
}
void ask_query(int k) {
    if(tr[k].l >= l && tr[k].r <= r) {ans = (ans + tr[k].w) % p; return;}
    if(tr[k].f) down(k);
    int mid = (tr[k].l + tr[k].r) >> 1;
    if(l <= mid) ask_query(lson);
    if(r > mid) ask_query(rson);
    tr[k].w = (tr[lson].w + tr[rson].w) % p;
}
void work1(int x, int y) {
    while(top[x] != top[y]) {
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        l = dfn[top[x]], r = dfn[x]; ask_query(1);
        x = fa[top[x]];
    }
    if(dep[x] > dep[y]) swap(x, y);
    l = dfn[x]; r = dfn[y]; ask_query(1);
}
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++) x = read(), y = read(), add(x, y), add(y, x);
    dfs(R); dfs1(R); build(1, 1, n);
    while(m--) {
        num = read();
        if(num == 1) cin >> x >> y >> z, work(x, y);
        if(num == 2) ans = 0, cin >> x >> y, work1(x, y), cout << ans << endl;
        if(num == 3) cin >> x >> z, l = dfn[x], r = dfn[x] + siz[x] - 1, change_query(1);
        if(num == 4) ans = 0, cin >> x, l = dfn[x], r = dfn[x] + siz[x] - 1, ask_query(1), cout << ans << endl;
    }
    return 0;
}

謝謝收看，祝身體健康！