UVa 11922 Permutation Transformer splay 把序列翻转后放到结尾

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18902

题意：给定原始序列 1-n ，m个操作，每次把区间[u,v]的数 翻转后放到序列结尾，最后中序遍历输出序列

第一发手打splay

#include <stdio.h>#include <string.h>#include <iostream>#include <queue>using namespace std;#define ll long long#define N 100105#define inf 100000000#define L(id) tree[id].ch[0]#define R(id) tree[id].ch[1]#define Father(id) tree[id].fall Mid(ll x, ll y){return (x+y)/2;}struct Node{ll ch[2], fa, size;ll num;bool filp;}tree[N];ll tot, root;void Newnode(ll& id, ll fa, ll val){id = ++tot;Father(id) = fa;tree[id].num = val;tree[id].size = 1;L(id) = R(id) = 0;tree[id].filp = 0;}void push_up(ll id){tree[id].size = tree[L(id)].size + tree[R(id)].size +1;}void push_down(ll id){if(tree[id].filp){tree[L(id)].filp ^= 1;tree[R(id)].filp ^= 1;swap(L(id), R(id));tree[id].filp = 0;}return ;}void Rotate(ll x, ll kind){// 0 代表左旋 1代表右旋 0:x在右子树 1:x在左子树ll y = Father(x);push_down(x); push_down(y);tree[y].ch[kind^1] = tree[x].ch[kind];Father(tree[x].ch[kind]) = y;if(Father(y))tree[Father(y)].ch[R(Father(y))==y] = x;Father(x) = Father(y);tree[x].ch[kind] = y;Father(y) = x;push_up(y);}void Splay(ll id, ll goal){push_down(id);while(Father(id)!=goal){ll y = Father(id);if(Father(y) == goal)Rotate(id, L(y)==id);else {ll kind = L(Father(y)) == y; //1表示y是左子树if(tree[y].ch[kind] == id){ //不共线Rotate(id, kind^1);Rotate(id, kind);}else {Rotate(y, kind);Rotate(id, kind);}}}push_up(id);if(goal == 0) root = id;}ll Get_kth(ll k){ //找到第k大的数int id = root;push_down(id);while(tree[L(id)].size!=k){if(tree[L(id)].size>k)id = L(id);else {k -= (tree[L(id)].size + 1);id = R(id);}push_down(id);}return id;}void RotateTo(ll l, ll r){Splay(Get_kth(l-1), 0);Splay(Get_kth(r+1), root);}void RotateMin(ll v, ll goal){int id = root;push_down(id);while(tree[id].num != v){if(tree[L(id)].num == v)id = L(id);else id = R(id);push_down(id);}Splay(id, goal);}void build(ll l, ll r, ll &id, ll fa){if(l > r)return;ll mid = Mid(l,r);Newnode(id, fa, mid);build(l, mid-1, L(id), id);build(mid+1, r, R(id), id);push_up(id);}ll getMax(ll id){push_down(id);while(R(id)){id = R(id);push_down(id);}return id;}int getMin(int id){push_down(id);while(L(id)){id = L(id);push_down(id);}return id;}void Print(ll id){push_down(id);if(L(id))Print(L(id));if(id != 1 && id != 2)printf("%lld\n",tree[id].num);if(R(id))Print(R(id));}ll n;void init(){tot = 0; root = 0;Father(0) = L(0) = R(0) = tree[0].size = 0;tree[0].num = -1;Newnode(root, 0, -1);Newnode(R(root), root, -1);build(1,n,L(R(root)),R(root));push_up(R(root));push_up(root);}int main(){ll que, u, v;while(~scanf("%lld %lld",&n,&que)){init();while(que--){scanf("%lld %lld", &u, &v);RotateTo(u, v);ll now = L(R(root));L(R(root)) = 0;push_up(R(root)); push_up(root);ll len = v-u+1;len = n - len +1;Splay(Get_kth(len),0);Splay(Get_kth(len-1), root);R(L(root)) = now;tree[now].fa = L(root);tree[now].filp ^= 1;push_up(L(root)); push_up(root);}Print(root);}return 0;}/*10 22 54 8题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=40758#problem/E*/