题解 P4008 【[NOI2003] 文本编辑器】
一道简单的数据结构题,直接用splay做即可
对于每一个操作
Move:直接修改光标位置
Insert:将光标所指的位置旋到根,再将它的后继旋到它下面,直接将序列插入为它的左儿子
Dlete:将光标所指的位置旋到根,再将光标加n所指的位置旋到它下面,并将它的左儿子直接赋值为0
Get:将光标所指的位置旋到根,再将光标加n所指的位置旋到它下面,输出它的左儿子
Prev:光标位置减1
Next:光标位置加1
代码
#include<iostream>
#include<cstdio>
using namespace std;
const int N = 2e6 + 50005;
struct tree
{
int s[2], v, p, siz;
void init(int _v, int _p)
{
s[0] = s[1] = 0;
v = _v, p = _p, siz = 1;
}//初始化
} tr[N];
int root, nodes[N], tot, poi;//nodes用来回收删除的节点
char s[N];
void pushup(int p)
{
tr[p].siz = tr[tr[p].s[0]].siz + tr[tr[p].s[1]].siz + 1;
}
int build(int l, int r, int p)
{
int u = nodes[tot--], mid = (l + r) >> 1;
tr[u].init((int)s[mid], p);
if (l < mid)
tr[u].s[0] = build(l, mid - 1, u);
if (r > mid)
tr[u].s[1] = build(mid + 1, r, u);
pushup(u);
return u;
}
void rotate(int x)
{
int y = tr[x].p, z = tr[y].p;
int k = tr[y].s[1] == x;
tr[z].s[tr[z].s[1] == y] = x, tr[x].p = z;
tr[y].s[k] = tr[x].s[k ^ 1], tr[tr[x].s[k ^ 1]].p = y;
tr[x].s[k ^ 1] = y, tr[y].p = x;
pushup(y), pushup(x);
}
void splay(int x, int k)
{
while (tr[x].p != k)
{
int y = tr[x].p, z = tr[y].p;
if (z != k)
{
if ((tr[y].s[1] == x) ^ (tr[z].s[1] == y))
rotate(x);
else
rotate(y);
}
rotate(x);
}
if (!k)
root = x;
}
int get_k(int k)
{
int u = root;
while (u)
{
if (tr[tr[u].s[0]].siz >= k)
u = tr[u].s[0];
else if (tr[tr[u].s[0]].siz + 1 == k)
return u;
else
k -= tr[tr[u].s[0]].siz + 1, u = tr[u].s[1];
}
return -1;
}
void dfs(int u)
{
if (tr[u].s[0])
dfs(tr[u].s[0]);
if (tr[u].s[1])
dfs(tr[u].s[1]);
nodes[++tot] = u;
}
void output(int u)
{
if (tr[u].s[0])
output(tr[u].s[0]);
printf("%c", (char)tr[u].v);
if (tr[u].s[1])
output(tr[u].s[1]);
}
int main()
{
for (int i = 1; i < N; i++)
nodes[++tot] = i;
int t;
scanf("%d", &t);
s[1] = 0, s[2] = 127;
root = build(1, 2, 0);
while (t--)
{
char op[10];
int k, n;
scanf("%s", &op);
if (op[0] == 'M')
{
scanf("%d", &k);
poi = k;
}
else if (op[0] == 'I')
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%c", &s[i]);
if ((int)s[i] > 126 || (int)s[i] < 32)
i--;
}
int l = get_k(poi + 1), r = get_k(poi + 2);
splay(l, 0), splay(r, l);
int v = build(1, n, r);
tr[r].s[0] = v;
pushup(r), pushup(l);
}
else if (op[0] == 'D')
{
scanf("%d", &n);
int l = get_k(poi + 1), r = get_k(poi + n + 2);
splay(l, 0), splay(r, l);
dfs(tr[r].s[0]);//回收节点
tr[r].s[0] = 0;
pushup(r), pushup(l);
}
else if (op[0] == 'G')
{
scanf("%d", &n);
int l = get_k(poi + 1), r = get_k(poi + n + 2);
splay(l, 0), splay(r, l);
output(tr[r].s[0]);
printf("\n");
}
else if (op[0] == 'P')
poi--;
else
poi++;
}
return 0;
}