[洛谷P3203][HNOI2010]弹飞绵羊
题目大意:有$n$个节点,第$i$个节点有一个弹力系数$k_i$,当到达第$i$个点时,会弹到第$i+k_i$个节点,若没有这个节点($i+k_i>n$)就会被弹飞。有两个操作:
- $x:$询问从第$x$个节点开始要多少次会被弹飞
- $x,y:$把第$x$个点的弹力系数修改为$y$。
题解:建一个节点标号$n+1$,所有大于$n$的位置都连向这,表示会被弹飞。其他每个节点向会被弹到的节点连边($i->k_i+i$),发现这样会构成一棵树。
询问就是问该点的深度(到$n+1$的距离)。修改就是把原来的边切断,再连上一条新边。可以想到$LCT$,只需要维护每个点到链顶的距离即可。
卡点:1.$get$函数写错
2.在$link$和$cut$操作中换了根,但是在询问中是按照根为$n+1$来做的
C++ Code:
#include <cstdio> #define maxn 200010 #define lc(rt) son[rt][0] #define rc(rt) son[rt][1] using namespace std; int sz[maxn], son[maxn][2], fa[maxn], tg[maxn]; inline int min(int a, int b) {return a < b ? a : b;} inline void swap(int &a, int &b) {a ^= b ^= a ^= b;} inline int get(int rt, int flag = 1) {return son[fa[rt]][flag] == rt;} inline bool is_root(int rt) {return !(get(rt, 0) || get(rt));} inline void update(int rt) {sz[rt] = sz[lc(rt)] + sz[rc(rt)] + 1;} inline void pushdown(int rt) { tg[rt] ^= 1; tg[lc(rt)] ^= 1; tg[rc(rt)] ^= 1; swap(lc(rt), rc(rt)); } inline void rotate(int x) { int y = fa[x], z = fa[y], b = get(x); if (!is_root(y)) son[z][get(y)] = x; fa[son[y][b] = son[x][!b]] = y; son[x][!b] = y; fa[y] = x, fa[x] = z; update(y), update(x); } int stack[maxn], top; inline void splay(int x) { stack[top = 1] = x; for (int y = x; !is_root(y); stack[++top] = y = fa[y]); for (; top; top--) if (tg[stack[top]]) pushdown(stack[top]); for (; !is_root(x); rotate(x)) if (!is_root(fa[x])) get(x) ^ get(fa[x]) ? rotate(x) : rotate(fa[x]); update(x); } inline void access(int rt) {for (int t = 0; rt; rc(rt) = t, t = rt, rt = fa[rt]) splay(rt);} inline void make_root(int rt) {access(rt), splay(rt), tg[rt] ^= 1;} inline void link(int x, int y) {make_root(x), fa[x] = y;} inline void split(int x, int y) {make_root(x), access(y), splay(y);} inline void cut(int x, int y) {split(x, y), lc(y) = fa[x] = 0;} int n, m; inline int ask(int rt) { make_root(n + 1); access(rt); splay(rt); return sz[rt] - 1; } int s[maxn]; int main() { scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d", &s[i]); link(i, min(n + 1, i + s[i])); } scanf("%d", &m); for (int i = 0; i < m; i++) { int op, x, y; scanf("%d%d", &op, &x); x++; if (op == 1) printf("%d\n", ask(x)); else { scanf("%d", &y); cut(x, min(n + 1, x + s[x])); s[x] = y; link(x, min(n + 1, x + y)); } } return 0; }