【BZOJ 3083】遥远的国度
这道题很简单的连剖+分类讨论,但是SDOI Round2要来了,不会手动栈怎么办呢?只好用一下这道题练习一下手动栈了,结果调了一天多QwQ
链剖的第一个dfs用bfs水过就行,但是我自以为是地把倍增写错了,坑了好久啊QAQ
这道题因为要询问子树,连剖的第二个dfs就不能再用bfs水过了,只能强行手动栈。
一开始拍小数据拍出了好多错,改过来后拍了无数组极限数据也拍不出来,因为构造的极限数据太弱了(偷懒构造u<v)根本无法暴露倍增的漏洞啊!!!
手残错误毁一生,写代码时不要自以为是。希望SD省选能够给我们一个更好的编程环境和评测环境(NOILinux恐怕是奢望吧)
#include<stack> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 100003; const int inf = 0x7fffffff; void read(int &k) { k = 0; int fh = 1; char c = getchar(); for(; c < '0' || c > '9'; c = getchar()) if (c == '-') fh = -1; for(; c >= '0' && c <= '9'; c = getchar()) k = (k << 1) + (k << 3) + c - '0'; k = k * fh; } struct node {int nxt, to;} E[N << 1]; int n, m, point[N], cnt = 0, pos[N], wt[N], deep[N]; int lazy[N * 8], mn[N * 8], top[N], sz[N], f[N][18], root, a[N], son[N]; void ins(int x, int y) {E[++cnt] = (node) {point[x], y}; point[x] = cnt;} stack <int> S; int qu[N]; void _() { int p = 0, q = 1; qu[1] = 1; while (p != q) { int u = qu[++p]; for(int tmp = point[u]; tmp; tmp = E[tmp].nxt) if (E[tmp].to != f[u][0]) f[E[tmp].to][0] = u, deep[E[tmp].to] = deep[u] + 1, qu[++q] = E[tmp].to; } for(int i = q; i >= 1; --i) { int u = qu[i], fa = f[u][0]; ++sz[u]; sz[fa] += sz[u]; if (sz[u] > sz[son[fa]]) son[fa] = u; } for(int j = 1; j <= 17; ++j) for(int i = 1; i <= n; ++i) f[i][j] = f[f[i][j - 1]][j - 1]; } void __() { top[1] = 1; cnt = 0; S.push(1); while (!S.empty()) { int u = S.top(); S.pop(); pos[++cnt] = u; wt[u] = cnt; for(int tmp = point[u]; tmp; tmp = E[tmp].nxt) if (E[tmp].to != f[u][0] && E[tmp].to != son[u]) top[E[tmp].to] = E[tmp].to, S.push(E[tmp].to); if (son[u]) {top[son[u]] = top[u]; S.push(son[u]);} } } /* 下面是正确的人工栈模板,完美模拟dfs,但是我学会它不到5小时我就用bfs把它淘汰了 void _() { S.push(Data(1, point[1])); sz[1] = 1; while (!S.empty()) { int x = S.top().x, y = S.top().y; S.pop(); if (y) { S.push(Data(x, E[y].nxt)); int v = E[y].to; if (v != f[x][0]) { f[v][0] = x; sz[v] = 1; deep[v] = deep[x] + 1; for(int i = 1; i <= 17; ++i) {f[v][i] = f[f[v][i - 1]][i - 1]; if (f[v][i] == 0) break;} S.push(Data(v, point[v])); } } else { if (!S.empty()) { sz[S.top().x] += sz[x]; if (sz[x] > sz[son[S.top().x]]) son[S.top().x] = x; } } } }*/ void pushdown(int rt) { if (lazy[rt]) { mn[rt << 1] = mn[rt << 1 | 1] = lazy[rt << 1] = lazy[rt << 1 | 1] = lazy[rt]; lazy[rt] = 0; } } void pushup(int rt) {mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);} void Build(int rt, int l, int r) { if (l == r) {mn[rt] = a[pos[l]]; mn[rt << 1] = mn[rt << 1 | 1] = inf; return;} int mid = (l + r) >> 1; Build(rt << 1, l, mid); Build(rt << 1 | 1, mid + 1, r); pushup(rt); } void add(int rt, int l, int r, int L, int R, int x) { if (L <= l && r <= R) {lazy[rt] = x; mn[rt] = x; return;} int mid = (l + r) >> 1; pushdown(rt); if (L <= mid) add(rt << 1, l, mid, L, R, x); if (R > mid) add(rt << 1 | 1, mid + 1, r, L, R, x); pushup(rt); } void Add(int x, int y, int z) { for(; top[x] != top[y]; x = f[top[x]][0]) { if (deep[top[x]] < deep[top[y]]) swap(x, y); add(1, 1, n, wt[top[x]], wt[x], z); } if (deep[x] < deep[y]) swap(x, y); add(1, 1, n, wt[y], wt[x], z); } int QQ(int rt, int l, int r, int L, int R) { if (L <= l && r <= R) return mn[rt]; int mid = (l + r) >> 1, s = inf; pushdown(rt); if (L <= mid) s = min(s, QQ(rt << 1, l, mid, L, R)); if (R > mid) s = min(s, QQ(rt << 1 | 1, mid + 1, r, L, R)); return s; } int Q(int L, int R) { if (L > n || R < 1 || L > R) return inf; return QQ(1, 1, n, L, R); } int lower; int LCA(int x, int y) { if (deep[x] < deep[y]) return 1; for(int i = 16; i >= 0; --i) if (deep[f[x][i]] > deep[y]) x = f[x][i]; lower = x; return f[x][0] != y; } int main() { read(n); read(m); int u, v, op, x; for(int i = 1; i < n; ++i) { read(u); read(v); ins(u, v); ins(v, u); } _(); __(); for(int i = 1; i <= n; ++i) read(a[i]); Build(1, 1, n); read(root); for(int i = 1; i <= m; ++i) { read(op); switch (op) { case 1: read(root); break; case 2: read(u); read(v); read(x); Add(u, v, x); break; case 3: read(u); if (root == u) printf("%d\n", Q(1, n)); else if (LCA(root, u)) { printf("%d\n", Q(wt[u], wt[u] + sz[u] - 1)); } else { v = min(Q(1, wt[lower] - 1), Q(wt[lower] + sz[lower], n)); printf("%d\n", v); } break; } } return 0; }
注释掉了能够完美模拟dfs但并没有什么用的手动栈= =
NOI 2017 Bless All