bzoj3626: [LNOI2014]LCA

题面在这里

做法：

首先说明以下都是以1为根的，点的编号是1~n。

考虑最暴力的做法：对于一个询问l, r, z，将1~z的点全部打标记，枚举l~r之间的点u，从1~u的路上找到第一个打标记的点，答案就加上这个点的深度。

然后稍微转化一下，假如把1~z的点权+1，然后对于l~r之间的点u，询问1~u之间的点权和，答案加上这个和。（正确性显然）

再转化，我们把l~r之间的点u，1~u上点权+1，然后最后询问1~z之间的点权和。这个和上面的操作是等价的。（显然）

以上的链加，链查询和都可以用树链剖分做到logn方。

然后对于所有的询问，我们考虑离线操作。

将所有询问排序后枚举1~n中点u，依次将1~u链上的点权+1，相应的询问中更新答案即可。

/*************************************************************Problem: bzoj 3626 [LNOI2014]LCAUser: fengyuanLanguage: C++Result: AcceptedTime: 1440 msMemory: 6188 kbSubmit_Time: 2017-12-05 22:58:13*************************************************************/#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<cctype>#include<vector>#include<map>#include<queue>#include<string>#define rep(i, x, y) for (int i = (x); i <= (y); i ++)#define down(i, x, y) for (int i = (x); i >= (y); i --)#define mid ((l+r)/2)#define lc (o<<1)#define rc (o<<1|1)#define pb push_back#define mp make_pair#define PII pair<int, int>#define F first#define S second#define B begin()#define E end()using namespace std;typedef long long LL;//headconst int N = 50010;const int MOD = 201314;int n, m, tot, cnt, clk;int head[N], z[N], ans[N], depth[N], fa[N], sz[N], son[N], top[N], in[N], sum[N<<2], tag[N<<2];struct Questions{int p, id, flag;}q[N<<1];struct Edge{int to, nex;}e[N];inline void add(int x, int y){e[++ cnt].to = y;e[cnt].nex = head[x];head[x] = cnt;}const bool cmp(const Questions &x, const Questions &y){return x.p < y.p;}inline void dfs(int u, int last, int s){depth[u] = s; fa[u] = last; sz[u] = 1;for (int i = head[u]; i; i = e[i].nex){int v = e[i].to; if (v == last) continue;dfs(v, u, s+1); sz[u] += sz[v];if (!son[u] || sz[v] > sz[son[u]]) son[u] = v;}}inline void dfs2(int u, int t){top[u] = t; in[u] = ++ clk;if (son[u]) dfs2(son[u], t);for (int i = head[u]; i; i = e[i].nex){int v = e[i].to; if (v == fa[u] || v == son[u]) continue;dfs2(v, v);}}inline void pushup(int o){sum[o] = sum[lc] + sum[rc];}inline void pushdown(int o, int l, int r){if (tag[o] == 0) return;tag[lc] += tag[o]; tag[rc] += tag[o];sum[lc] += tag[o]*(mid-l+1); sum[rc] += tag[o]*(r-mid);tag[o] = 0;}inline void update(int o, int l, int r, int x, int y){if (l == x && r == y){sum[o] += r-l+1; tag[o] ++;return;}pushdown(o, l, r);if (y <= mid) update(lc, l, mid, x, y);else if (x > mid) update(rc, mid+1, r, x, y);else update(lc, l, mid, x, mid), update(rc, mid+1, r, mid+1, y);pushup(o);}inline void change(int x, int y){while (top[x] != top[y]){if (depth[top[x]] < depth[top[y]]) swap(x, y);update(1, 1, n, in[top[x]], in[x]);x = fa[top[x]];}if (depth[x] > depth[y]) swap(x, y);update(1, 1, n, in[x], in[y]);}inline int query(int o, int l, int r, int x, int y){if (l == x && r == y) return sum[o];pushdown(o, l, r);if (y <= mid) return query(lc, l, mid, x, y);else if (x > mid) return query(rc, mid+1, r, x, y);else return query(lc, l, mid, x, mid) + query(rc, mid+1, r, mid+1, y);}inline int solve(int x, int y){int ret = 0;while (top[x] != top[y]){if (depth[top[x]] < depth[top[y]]) swap(x, y);ret = (ret+query(1, 1, n, in[top[x]], in[x]))%MOD;x = fa[top[x]];}if (depth[x] > depth[y]) swap(x, y);return (ret+query(1, 1, n, in[x], in[y]))%MOD;}int main(){scanf("%d%d", &n, &m);rep(i, 2, n){int x; scanf("%d", &x); x ++;add(x, i);}dfs(1, 0, 0); dfs2(1, 1);rep(i, 1, m){int x, y; scanf("%d%d%d", &x, &y, &z[i]);x ++, y ++, z[i] ++;q[++ tot].p = x-1; q[tot].id = i; q[tot].flag = 0;q[++ tot].p = y; q[tot].id = i; q[tot].flag = 1;}int now = 0;sort(q+1, q+1+tot, cmp);rep(i, 1, tot){while (now < q[i].p) change(1, ++ now);int t = z[q[i].id];if (!q[i].flag) ans[q[i].id] = (ans[q[i].id]-solve(1, t)+MOD)%MOD;else ans[q[i].id] = (ans[q[i].id]+solve(1, t))%MOD;}rep(i, 1, m) printf("%d\n", ans[i]);return 0;}