标签:printf pos tree 标记 argv max public 满二叉树 read
二次联通门 : luogu P3391 文艺平衡树
/* luogu 3391 文艺平衡树 splay 区间翻转 每次翻转区间[l,r]都是把l旋到根节点上, r旋到根节点的右节点上 那么要修改的区间就是根节点的右孩子的左子树 然后打标记。。 之后每次查到时就下放标记 原区间为1 ~ N 增加两个哨兵节点 0 和 N + 1, 避免翻转1~N的区间产生麻烦 */ #include <cstdio> #define Max 200009 inline int swap (int &a, int &b) { int now = a; a = b; b = now; } inline void read (int &now) { now = 0; register char word = getchar (); while (word < ‘0‘ || word > ‘9‘) word = getchar (); while (word >= ‘0‘ && word <= ‘9‘) { now = now * 10 + word - ‘0‘; word = getchar (); } } int value[Max]; int N, M; class Splay_Tree_Type { private : struct Splay_Tree_Date { int size; int key; int father; int child[2]; int Flandre; } tree[Max]; inline int Get_Son (int now) { return tree[tree[now].father].child[1] == now; } inline void Update (int now) { tree[now].size = 1; if (tree[now].child[0]) tree[now].size += tree[tree[now].child[0]].size; if (tree[now].child[1]) tree[now].size += tree[tree[now].child[1]].size; } int Root; int Answer[Max]; int Count; inline void Down (int now) { tree[now].Flandre = 0; swap (tree[now].child[0], tree[now].child[1]); if (tree[now].child[0]) tree[tree[now].child[0]].Flandre ^= 1; if (tree[now].child[1]) tree[tree[now].child[1]].Flandre ^= 1; } inline void Rotate (int now) { int father = tree[now].father; int Grand = tree[father].father; int pos = Get_Son (now); if (tree[father].Flandre && father) // 注意这里标记下放的次序 Down (father); if (tree[now].Flandre && now) Down (now); tree[father].child[pos] = tree[now].child[pos ^ 1]; tree[tree[father].child[pos]].father = father; tree[now].child[pos ^ 1] = father; tree[father].father = now; tree[now].father = Grand; if (Grand) tree[Grand].child[tree[Grand].child[1] == father] = now; Update (father); Update (now); } void Get_Answer (int now) // 把整棵树进行中序遍历后的序列就是答案 { if (tree[now].Flandre) Down (now); if (tree[now].child[0]) Get_Answer (tree[now].child[0]); Answer[++Count] = tree[now].key; if (tree[now].child[1]) Get_Answer (tree[now].child[1]); } int Build (int l, int r, int father) { int now = l + r >> 1; tree[now].father = father; tree[now].key = value[now]; if (now > l) tree[now].child[0] = Build (l, now - 1, now); if (now < r) tree[now].child[1] = Build (now + 1, r, now); Update (now); return now; } public : void Prepare () { Root = 1; Root = Build(1, N + 2, 0); } void Splay (int now, int to) { for (int father; (father = tree[now].father) != to; Rotate (now)) if (tree[father].father != to) Rotate (Get_Son (now) == Get_Son (father) ? father : now); if (!to) Root = now; } int Get_Pos (int x) { int now = Root; for (; ; ) { if (tree[now].Flandre) Down (now); if (x <= tree[tree[now].child[0]].size) now = tree[now].child[0]; else { x -= tree[tree[now].child[0]].size + 1; if (!x) return now; now = tree[now].child[1]; } } } inline void Print () { Get_Answer (Root); for (int i = 1; i <= N; i++) printf ("%d ", Answer[i + 1]); } inline void Hit_flag () { tree[tree[tree[Root].child[1]].child[0]].Flandre ^= 1; } }; Splay_Tree_Type Make; int main (int argc, char *argv[]) { read (N); read (M); // 0号点和 N+1 号点代表是哨兵节点 for (int i = 1; i <= N + 2; i++) // 预先处理出值 value[i] = i - 1; int l, r; Make.Prepare (); // splay建树, 要建满二叉树 for (; M--; ) { read (l); read (r); r += 2; l = Make.Get_Pos (l); // 找到在树中对应的位置 r = Make.Get_Pos (r); Make.Splay (l, 0); Make.Splay (r, l); Make.Hit_flag (); // 打上标记 } Make.Print (); return 0; }
标签:printf pos tree 标记 argv max public 满二叉树 read
原文地址:http://www.cnblogs.com/ZlycerQan/p/6882365.html