hdu3436Queue-jumpers(splay+离散化)
题目请戳这里
题目大意：有一个队，n个人，编号1-n，有3种操作：
top x：将编号x的人放到队列首；
query x：查询编号x的人现在排在第几位；
rank x：查询现在第x位的人编号；
题目分析：splay。人数编号是10^8级别的，但是操作不超过10^5个，只需将top操作中的编号离散化就可以了，因为剩下的人都是连续的，所以将剩下的连续的人缩成一个点，只记录起始位置的人和连续长度即可。
trick：n可以为1，这种情况下，任何操作都是无意义的，遇到查询操作就输出1。。这点TLE出翔了。。。
感想：splay非常强大，但是如果不能1y，那就意味着悲剧。。。
详情请见代码：
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 200005;
typedef __int64 ll;
int hash[N];
int add[N];
int index[N];
int next[N];
struct spt
{
int l,r,f,val,size,len;
}tree[N];
int n,m,q;
struct node
{
char cmd;
int x;
}op[N];
void init()
{
for(int i = 0;i < N-10;i ++)
next[i] = i + 1;
}
int newnode(int st,int ed)
{
int p = next[0];
next[0] = next[p];
tree[p].l = tree[p].r = tree[p].f = 0;
tree[p].val = st;
tree[p].size = tree[p].len = ed - st;
return p;
}
void delnode(int p)
{
next[p] = next[0];
next[0] = p;
}
void del(int rt)
{
if(!rt)
return;
del(tree[rt].l);
del(tree[rt].r);
delnode(rt);
}
void pushup(int rt)
{
if(!rt)
return;
tree[rt].size = tree[tree[rt].l].size + tree[tree[rt].r].size + tree[rt].len;
}
void zig(int x)
{
int p = tree[x].f;
tree[p].l = tree[x].r;
if(tree[x].r)
tree[tree[x].r].f = p;
pushup(p);
tree[x].r = p;
tree[x].f = tree[p].f;
pushup(x);
tree[p].f = x;
if(tree[x].f == 0)
return;
if(tree[tree[x].f].l == tree[x].r)
tree[tree[x].f].l = x;
else
tree[tree[x].f].r = x;
}
void zag(int x)
{
int p = tree[x].f;
tree[p].r = tree[x].l;
if(tree[x].l)
tree[tree[x].l].f = p;
pushup(p);
tree[x].l = p;
tree[x].f = tree[p].f;
pushup(x);
tree[p].f = x;
if(tree[x].f == 0)
return;
if(tree[tree[x].f].l == tree[x].l)
tree[tree[x].f].l = x;
else
tree[tree[x].f].r = x;
}
int splay(int x,int goal)
{
while(tree[x].f != goal)
{
int p = tree[x].f;
int g = tree[p].f;
if(g == goal)
{
if(tree[p].l == x)
zig(x);
if(tree[p].r == x)
zag(x);
}
else
{
if(tree[g].l == p && tree[p].l == x)
zig(p),zig(x);
else if(tree[g].l == p && tree[p].r == x)
zag(x),zig(x);
else if(tree[g].r == p && tree[p].l == x)
zig(x),zag(x);
else if(tree[g].r == p && tree[p].r == x)
zag(p),zag(x);
}
}
pushup(x);
return x;
}
int build(int l,int r,int f)
{
if(l > r)
return 0;
int mid = (l + r)>>1;
int p = newnode(hash[mid],hash[mid + 1]);
tree[p].l = build(l,mid - 1,p);
tree[p].f = f;
tree[p].r = build(mid + 1,r,p);
pushup(p);
add[mid] = p;
return p;
}
int getfront(int rt)
{
while(tree[rt].l)
{
rt = tree[rt].l;
}
return rt;
}
int getval(int pos,int rt)
{
if(!rt)
return 0;
if(pos <= tree[tree[rt].l].size)
return getval(pos,tree[rt].l);
if(pos <= tree[tree[rt].l].size + tree[rt].len)
return rt;//pos - tree[tree[rt].l].size - 1 + tree[rt].val;
return getval(pos - tree[tree[rt].l].size - tree[rt].len,tree[rt].r);
}
int Bin(int x)
{
int l,r,mid;
l = 1;r = m;
while(r >= l)
{
mid = (l + r)>>1;
if(hash[mid] < x && hash[mid + 1] > x)
return mid;
if(hash[mid - 1] < x && hash[mid] > x)
return mid - 1;
if(hash[mid] == x)
return mid;
else if(hash[mid] > x)
r = mid - 1;
else
l = mid + 1;
}
}
void Top(int x,int &root)
{
int p = Bin(x);
root = splay(add[p],0);
if(tree[root].l == 0)
{
root = tree[root].r;
tree[root].f = 0;
}
else if(tree[root].r == 0)
{
root = tree[root].l;
tree[root].f = 0;
}
else
{//后继旋到右子树，左子树挂到右子树，删除根
int nt = getfront(tree[root].r);
tree[root].r = splay(nt,root);
tree[tree[root].l].f = tree[root].r;
tree[tree[root].r].l = tree[root].l;
tree[tree[root].r].f = 0;
root = tree[root].r;
}
pushup(root);
int tmp = getfront(root);
root = splay(tmp,0);
tree[root].l = add[p];
tree[add[p]].f = root;
tree[add[p]].l = tree[add[p]].r = 0;
root = splay(add[p],0);
}
void Query(int x,int &root)
{
int p = Bin(x);
root = splay(add[p],0);
int ans = tree[tree[root].l].size + x - tree[root].val + 1;
printf("%d\n",ans);
}
void Rank(int x,int &root)
{
int p = getval(x,root);
root = splay(p,0);
printf("%d\n",x - tree[tree[root].l].size - 1 + tree[root].val);
}
int main()
{
int _,i,j;
int cas = 0;
char ss[10];
int root;
init();
scanf("%d",&_);
while(_--)
{
scanf("%d%d",&n,&q);
printf("n:%d q:%d\n",n,q);
j = 1;
printf("Case %d:\n",++cas);
if(n == 1)
{
for(i = 1;i <= q;i ++)
{
scanf("%s%d",ss,&j);
if(*ss == 'R' || *ss == 'Q')
printf("1\n");
}
continue;
}
for(i = 1;i <= q;i ++)
{
scanf("%s",ss);
op[i].cmd = *ss;
scanf("%d",&op[i].x);
if(*ss == 'T')
{
hash[j] = op[i].x;
j ++;
}
}
sort(hash + 1,hash + j);
m = 2;
for(i = 2;i < j;i ++)
if(hash[i] != hash[i - 1])
hash[m ++] = hash[i];
for(i = 2,j = m;i < j;i ++)
if(hash[i] != hash[i - 1] + 1)
hash[m ++] = hash[i - 1] + 1;
if(hash[j - 1] < n)
hash[m ++] = hash[j - 1] + 1;
hash[m ++] = n + 1;
if(hash[1] != 1)
hash[m ++] = 1;
sort(hash + 1,hash + m);
m --;
root = build(1,m - 1,0);
for(i = 1;i <= q;i ++)
{
switch(op[i].cmd)
{
case 'T':Top(op[i].x,root);break;
case 'Q':Query(op[i].x,root);break;
case 'R':Rank(op[i].x,root);break;
}
}
del(root);
}
return 0;
}
//140MS 4356K