UVA_11994
这个题目思维上的障碍比较少,因为实际上都是link-cut-tree的基本的操作,还有一个更为复杂的link-cut-tree的题目——HDU_4010。
在维护路径上边的颜色的数量时,由于颜色的种类只有30个,因此可以用一个整数的30个二进制位来存储颜色的种类。
#include<stdio.h> #include<string.h> #define MAXD 50010 int N, M; struct Splay { int pre, ls, rs, col, set, to, size; bool root; void pushdown(); void update(); void zig(int x); void zag(int x); void splay(int x); void init() { root = true; size = to = pre = ls = rs = col = set = 0; } }sp[MAXD]; inline void same(int x, int c) { if(x) sp[x].to = sp[x].col = c, sp[x].set = 1 << c; } inline void Splay::pushdown() { if(to) { same(ls, to), same(rs, to); to = 0; } } inline void Splay::update() { set = sp[ls].set | sp[rs].set | 1 << col; size = sp[ls].size + sp[rs].size + 1; } void Splay::zig(int x) { int y = rs, fa = pre; rs = sp[y].ls, sp[rs].pre = x; sp[y].ls = x, pre = y; sp[y].pre = fa; if(root) root = false, sp[y].root = true; else sp[fa].rs == x ? sp[fa].rs = y : sp[fa].ls = y; update(); } void Splay::zag(int x) { int y = ls, fa = pre; ls = sp[y].rs, sp[ls].pre = x; sp[y].rs = x, pre = y; sp[y].pre = fa; if(root) root = false, sp[y].root = true; else sp[fa].rs == x ? sp[fa].rs = y : sp[fa].ls = y; update(); } void Splay::splay(int x) { int y, z; for(pushdown(); !root;) { y = pre; if(sp[y].root) sp[y].pushdown(), pushdown(), sp[y].rs == x ? sp[y].zig(y) : sp[y].zag(y); else { z = sp[y].pre, sp[z].pushdown(), sp[y].pushdown(), pushdown(); if(sp[z].rs == y) { if(sp[y].rs == x) sp[z].zig(z), sp[y].zig(y); else sp[y].zag(y), sp[z].zig(z); } else { if(sp[y].ls == x) sp[z].zag(z), sp[y].zag(y); else sp[y].zig(y), sp[z].zag(z); } } } update(); } void access(int x) { int fx; for(fx = x, x = 0; fx != 0; x = fx, fx = sp[x].pre) { sp[fx].splay(fx); sp[sp[fx].rs].root = true, sp[fx].rs = x, sp[x].root = false; sp[fx].update(); } } void cut(int x) { access(x), sp[x].splay(x); sp[sp[x].ls].root = true, sp[sp[x].ls].pre = 0, sp[x].ls = 0; sp[x].update(); } void join(int x, int y, int c) { access(y), sp[y].splay(y); sp[x].splay(x); if(sp[y].pre != 0) return ; cut(x); sp[x].pre = y, sp[x].col = c; sp[x].update(); } void init() { int i, x; for(i = 1; i <= N; i ++) { scanf("%d", &x); sp[i].init(), sp[i].pre = x; } for(i = 1; i <= N; i ++) { scanf("%d", &x); sp[i].col = x, sp[i].set = 1 << x, sp[i].size = 1; } sp[0].init(); } void paint(int x, int y, int c) { int fx; access(y), sp[y].splay(y); for(fx = x, x = 0; fx != 0; x = fx, fx = sp[x].pre) { sp[fx].splay(fx); if(sp[fx].pre == 0) { if(sp[y].pre != 0 || fx == y) same(sp[fx].rs, c), same(x, c); } sp[sp[fx].rs].root = true, sp[fx].rs = x, sp[x].root = false; sp[fx].update(); } } void count(int x, int y) { int fx; access(y), sp[y].splay(y); for(fx = x, x = 0; fx != 0; x = fx, fx = sp[x].pre) { sp[fx].splay(fx); if(sp[fx].pre == 0) { if(sp[y].pre == 0 && fx != y) printf("0 0\n"); else { int cnt = 0, t = sp[sp[fx].rs].set | sp[x].set; for(int i = 1; i <= 30; i ++) if(t & 1 << i) ++ cnt; printf("%d %d\n", sp[sp[fx].rs].size + sp[x].size, cnt); } } sp[sp[fx].rs].root = true, sp[fx].rs = x, sp[x].root = false; sp[fx].update(); } } void solve() { int i, op, x, y, c; for(i = 0; i < M; i ++) { scanf("%d", &op); if(op == 1) { scanf("%d%d%d", &x, &y, &c); if(x == y) continue; join(x, y, c); } else if(op == 2) { scanf("%d%d%d", &x, &y, &c); if(x == y) continue; paint(x, y, c); } else { scanf("%d%d", &x, &y); if(x == y) printf("0 0\n"); else count(x, y); } } } int main() { while(scanf("%d%d", &N, &M) == 2) { init(); solve(); } return 0; }