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【luogu P3384 树链剖分】 模板

發布時間:2024/9/5 编程问答 37 豆豆
生活随笔 收集整理的這篇文章主要介紹了 【luogu P3384 树链剖分】 模板 小編覺得挺不錯的,現在分享給大家,幫大家做個參考.

題目鏈接:https://www.luogu.org/problemnew/show/P3384
誒又給自己留了個坑..不想寫線段樹一大理由之前的模板變量名太長

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define lson left, mid, rt<<1 #define rson mid + 1, right, rt<<1|1 #define ll long long using namespace std; const int maxn = 200000 + 10; ll n, m, root, mod; ll node[maxn], num; ll fa[maxn], dep[maxn], size[maxn], son[maxn], top[maxn], seg[maxn], rev[maxn]; ll res; struct edge{ll from, to, next; }e[maxn<<2]; ll head[maxn], cnt; //------------------------------------------------------- ll tree[maxn<<2], lazy[maxn<<2]; void PushUP(ll rt) {tree[rt] = (tree[rt<<1] + tree[rt<<1|1])%mod; } void build(ll left, ll right, ll rt) {if(left == right){tree[rt] = rev[left];tree[rt] = tree[rt]%mod;return;}ll mid = (left + right) >> 1;build(lson);build(rson);PushUP(rt); } void PushDOWN(ll left, ll right, ll rt, ll mid) {lazy[rt<<1] += lazy[rt]; lazy[rt<<1|1] += lazy[rt];tree[rt<<1] += ((mid - left + 1)*lazy[rt])%mod;tree[rt<<1|1] += ((right - mid)*lazy[rt])%mod;lazy[rt] = 0; } ll query(ll l, ll r, ll left, ll right, ll rt) {ll res = 0;if(l <= left && r >= right){return tree[rt]%mod;}ll mid = (left + right)>>1;if(lazy[rt]) PushDOWN(left, right, rt, mid);if(l <= mid) res += query(l, r, lson);if(r > mid) res += query(l, r, rson);return res; } void update(ll l, ll r, ll add, ll left, ll right, ll rt) {if(l <= left && r >= right){lazy[rt] += add;tree[rt] += (right - left + 1)*add;return;}ll mid = (left + right)>>1;PushDOWN(left, right, rt, mid);if(l <= mid) update(l, r, add, lson);if(r > mid) update(l, r, add, rson);PushUP(rt); } //------------------------------------------------ void add(ll u, ll v) {e[++cnt].from = u;e[cnt].next = head[u];e[cnt].to = v;head[u] = cnt; } void dfs1(ll u, ll f, ll d) {ll maxson = -1;size[u] = 1;fa[u] = f;dep[u] = d;for(ll i = head[u]; i != -1; i = e[i].next){ll v = e[i].to;if(v != f){dfs1(v, u, d + 1);size[u] += size[v];if(size[v] > maxson) son[u] = v, maxson = size[v];}} } void dfs2(ll u, ll t) {seg[u] = ++num;rev[num] = node[u];top[u] = t;if(!son[u]) return;dfs2(son[u], t);for(ll i = head[u]; i != -1; i = e[i].next){ll v = e[i].to;if(fa[u] == v || son[u] == v) continue;dfs2(v, v);} } ll qRange(ll x, ll y) {ll ans = 0;while(top[x] != top[y]){if(dep[top[x]] < dep[top[y]]) swap(x, y);res = 0;res = query(seg[top[x]], seg[x], 1, n, 1);ans = (ans + res)%mod;x = fa[top[x]];}if(dep[x] > dep[y]) swap(x, y);res = 0;res = query(seg[x], seg[y], 1, n, 1);ans = (ans + res)%mod;return ans; } void updRange(ll x, ll y, ll k) {k = k%mod;while(top[x] != top[y]){if(dep[top[x]] < dep[top[y]]) swap(x, y);update(seg[top[x]], seg[x], k, 1, n, 1);x = fa[top[x]];}if(dep[x] > dep[y]) swap(x, y);update(seg[x], seg[y], k, 1, n, 1); } ll qSon(ll x) {res = 0;res = query(seg[x], seg[x]+size[x]-1, 1, n, 1);return res; } ll updSon(ll x, ll k) {update(seg[x], seg[x]+size[x]-1, k, 1, n, 1); } int main() {memset(head, -1, sizeof(head));scanf("%lld%lld%lld%lld",&n,&m,&root,&mod);for(ll i = 1; i <= n; i++)scanf("%lld",&node[i]);for(ll i = 1; i < n; i++){ll u, v;scanf("%lld%lld",&u,&v);add(u, v), add(v, u);}dfs1(root, 0, 1);dfs2(root, root);build(1,n,1);for(ll i = 1; i <= m; i++){ll opt, x, y, z;scanf("%lld",&opt);if(opt == 1){scanf("%lld%lld%lld",&x,&y,&z);updRange(x, y, z);}if(opt == 2){scanf("%lld%lld",&x,&y);printf("%lld\n",qRange(x, y)%mod);}if(opt == 3){scanf("%lld%lld",&x,&y);updSon(x, y);}if(opt == 4){scanf("%lld",&x);printf("%lld\n",qSon(x)%mod);}}return 0; }

轉載于:https://www.cnblogs.com/MisakaAzusa/p/9409727.html

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