传送门:HDU-4453
题解:Splay树
如果要把指针从1移到2,可以把1先删除,再插入到n的后面,如果要把指针从1移动到n,可以把n删除,再插入到1的前面
#include<bits/stdc++.h>
using namespace std;
const int MX = 2e5 + 5;
int m, k1, k2;
int n;
int a[MX];
int root, rear; //根节点,节点总数
int rem[MX], tot; //经过删除后未被使用的节点
int ch[MX][2], fa[MX];
int val[MX], col[MX], add[MX];
int sz[MX];
void NewNode(int &rt, int father, int v) {
if (tot) rt = rem[tot--];
else rt = ++rear;
fa[rt] = father;
ch[rt][0] = ch[rt][1] = col[rt] = add[rt] = 0;
val[rt] = v;
sz[rt] = 1;
}
void PushUP(int rt) {
int ls = ch[rt][0], rs = ch[rt][1];
sz[rt] = sz[ls] + sz[rs] + 1;
}
void PushDown(int rt) {
if (col[rt]) {
col[ch[rt][0]] ^= col[rt];
col[ch[rt][1]] ^= col[rt];
swap(ch[rt][0], ch[rt][1]);
col[rt] = 0;
}
if (add[rt]) {
add[ch[rt][0]] += add[rt];
add[ch[rt][1]] += add[rt];
val[ch[rt][0]] += add[rt];
val[ch[rt][1]] += add[rt];
add[rt] = 0;
}
}
void build(int &rt, int l, int r, int father) {
if (l > r) return;
int m = (l + r) >> 1;
NewNode(rt, father, a[m]);
build(ch[rt][0], l, m - 1, rt);
build(ch[rt][1], m + 1, r, rt);
PushUP(rt);
}
void init() {
root = rear = tot = 0;
NewNode(root, 0, 0); //一共n+2个节点,0~n+1
NewNode(ch[root][1], root, 0);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
build(ch[ch[root][1]][0], 1, n, ch[root][1]);
PushUP(root);
}
void Link(int x, int y, int c) {
fa[x] = y; ch[y][c] = x;
}
void Rotate(int x, int c) { //c=0表示左旋,c=1表示右旋
int y = fa[x];
PushDown(y);
PushDown(x);
Link(x, fa[y], ch[fa[y]][1] == y);
Link(ch[x][c], y, !c);
Link(y, x, c);
PushUP(y); //y变成x子节点,只要更新y
}
void Splay(int x, int f) {
PushDown(x);
while (fa[x] != f) {
int y = fa[x];
if (fa[y] == f) Rotate(x, ch[y][0] == x);
else {
int t = ch[fa[y]][0] == y;
if (ch[y][t] == x) Rotate(x, !t);
else Rotate(y, t);
Rotate(x, t);
}
}
PushUP(x); //更新x
if (f == 0) root = x;
}
int get_kth(int rt, int k) {
PushDown(rt);
int t = sz[ch[rt][0]] + 1, ret;
if (t == k) ret = rt;
else if (t > k) ret = get_kth(ch[rt][0], k);
else ret = get_kth(ch[rt][1], k - t);
PushUP(rt);
return ret;
}
void Update(int L, int R, int v) {
Splay(get_kth(root, L), 0);
Splay(get_kth(root, R + 2), root);
int t = ch[ch[root][1]][0];
add[t] += v;
val[t] += v;
}
void Reverse(int L, int R) {
Splay(get_kth(root, L), 0);
Splay(get_kth(root, R + 2), root);
int t = ch[ch[root][1]][0];
col[t] ^= 1;
}
void Insert(int p, int v) {
Splay(get_kth(root, p + 1), 0); //将第p个数旋转到根节点
Splay(get_kth(root, p + 2), root); //将第p+1个数旋转到根节点的右儿子,此时该节点没有左儿子
NewNode(ch[ch[root][1]][0], ch[root][1], v);
PushUP(ch[root][1]);
PushUP(root);
n++;
}
void erase(int rt) {
if (!rt)return;
fa[rt] = 0;
rem[++tot] = rt;
n--;
erase(ch[rt][0]);
erase(ch[rt][1]);
}
int Delete(int p) {
Splay(get_kth(root, p), 0); //将第p-1个数旋转到根节点
Splay(get_kth(root, p + 2), root);//将第p+1个数旋转到根节点的右儿子,此时该节点左儿子为p
int t = ch[ch[root][1]][0];
erase(t);
ch[ch[root][1]][0] = 0;
PushUP(ch[root][1]);
PushUP(root);
return t;
}
int Query(int p) {
Splay(get_kth(root, p + 1), 0); //将第p个数旋转到根节点
return val[root];
}
void Move(int x) {
if (x == 1) {
int t = Delete(n);
Insert(0, val[t]);
} else {
int t = Delete(1);
Insert(n, val[t]);
}
}
int main() {
char op[10];
int x, cas = 0;
//freopen("in.txt","r",stdin);
while (scanf("%d%d%d%d", &n, &m, &k1, &k2), n || m || k1 || k2) {
printf("Case #%d:\n", ++cas);
init();
while (m--) {
scanf("%s", op);
if (op[0] == 'a') {
scanf("%d", &x);
Update(1, k2, x);
} else if (op[0] == 'r') Reverse(1, k1);
else if (op[0] == 'i') {
scanf("%d", &x);
Insert(1, x);
} else if (op[0] == 'd') {
Delete(1);
} else if (op[0] == 'm') {
scanf("%d", &x);
Move(x);
} else printf("%d\n", Query(1));
}
}
return 0;
}