首页 > 代码库 > POJ 3667 splay区间合并练习
POJ 3667 splay区间合并练习
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 12446 | Accepted: 5363 |
Description
The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny shores of Lake Superior. Bessie, ever the competent travel agent, has named the Bullmoose Hotel on famed Cumberland Street as their vacation residence. This immense hotel has N (1 ≤ N ≤ 50,000) rooms all located on the same side of an extremely long hallway (all the better to see the lake, of course).
The cows and other visitors arrive in groups of size Di (1 ≤Di ≤ N) and approach the front desk to check in. Each group i requests a set of Di contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbersr..r+Di-1 if they are available or, if no contiguous set of rooms is available, politely suggests alternate lodging. Canmuu always chooses the value ofr to be the smallest possible.
Visitors also depart the hotel from groups of contiguous rooms. Checkout i has the parameters Xi andDi which specify the vacating of rooms Xi ..Xi +Di-1 (1 ≤Xi ≤ N-Di+1). Some (or all) of those rooms might be empty before the checkout.
Your job is to assist Canmuu by processing M (1 ≤ M < 50,000) checkin/checkout requests. The hotel is initially unoccupied.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Line i+1 contains request expressed as one of two possible formats: (a) Two space separated integers representing a check-in request: 1 andDi (b) Three space-separated integers representing a check-out: 2,Xi, and Di
Output
* Lines 1.....: For each check-in request, output a single line with a single integerr, the first room in the contiguous sequence of rooms to be occupied. If the request cannot be satisfied, output 0.
Sample Input
10 6 1 3 1 3 1 3 1 3 2 5 5 1 6
Sample Output
1 4 7 0 5
代码：
/* ***********************************************
Author :rabbit
Created Time :2014/11/1 12:57:54
File Name :4.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=200200;
struct Node;
Node *null;
struct Node{
Node *ch[2],*fa;
int size;
int lsum,rsum,msum,col,val;
Node(){
ch[0]=ch[1]=fa=null;
col=0;
}
inline void setc(Node *p,int d){
ch[d]=p;
p->fa=this;
}
inline bool d(){
return fa->ch[1]==this;
}
void clear(){
size=1;
ch[0]=ch[1]=fa=null;
lsum=rsum=msum=val=1;
col=-1;
}
inline void push_up(){
if(this==null)return;
size=ch[0]->size+ch[1]->size+1;
lsum=ch[0]->lsum;
rsum=ch[1]->rsum;
msum=max(ch[0]->msum,ch[1]->msum);
if(val){
msum=max(msum,ch[0]->rsum+ch[1]->lsum+1);
if(lsum==ch[0]->size)lsum+=ch[1]->lsum+1;
if(rsum==ch[1]->size)rsum+=ch[0]->rsum+1;
}
}
void update(int c){
if(this==null)return;
lsum=rsum=msum=c*size;
val=col=c;
}
inline void push_down(){
if(this==null)return;
if(col!=-1){
ch[0]->update(col);
ch[1]->update(col);
col=-1;
}
}
};
inline void rotate(Node *x){
Node *f=x->fa,*ff=x->fa->fa;
f->push_down();
x->push_down();
int c=x->d(),cc=f->d();
f->setc(x->ch[!c],c);
x->setc(f,!c);
if(ff->ch[cc]==f)ff->setc(x,cc);
else x->fa=ff;
f->push_up();
}
inline void splay(Node *&root,Node *x,Node *goal){
while(x->fa!=goal){
if(x->fa->fa==goal)rotate(x);
else{
x->fa->fa->push_down();
x->fa->push_down();
x->push_down();
bool f=x->fa->d();
x->d()==f?rotate(x->fa):rotate(x);
rotate(x);
}
}
x->push_up();
if(goal==null)root=x;
}
Node *get_kth(Node *r,int k){
Node *x=r;
x->push_down();
while(x->ch[0]->size+1!=k){
if(k<x->ch[0]->size+1)x=x->ch[0];
else{
k-=x->ch[0]->size+1;
x=x->ch[1];
}
x->push_down();
}
return x;
}
Node pool[maxn],*tail,*node[maxn],*root;
void build(Node *&x,int l,int r,Node *fa){
if(l>r)return;
int mid=(l+r)/2;
x=tail++;
x->clear();
x->fa=fa;
node[mid]=x;
build(x->ch[0],l,mid-1,x);
build(x->ch[1],mid+1,r,x);
x->push_up();
}
void init(int n){
tail=pool;
null=tail++;
null->fa=null->ch[0]=null->ch[1]=null;
null->size=0;
null->val=null->lsum=null->rsum=null->msum=0;null->col=-1;
Node *p=tail++;
p->val=p->msum=p->rsum=p->lsum=0;p->col=-1;
p->size=1;p->ch[0]=p->ch[1]=p->fa=null;
root=p;
p=tail++;
p->val=p->msum=p->rsum=p->lsum=0;p->col=-1;
p->size=1;p->ch[0]=p->ch[1]=p->fa=null;
root->setc(p,1);
build(root->ch[1]->ch[0],1,n,root->ch[1]);
root->ch[1]->push_up();
root->push_up();
}
void update(int l,int r,int c){
splay(root,get_kth(root,l),null);
splay(root,get_kth(root,r+2),root);
root->ch[1]->ch[0]->update(c);
root->ch[1]->push_up();
root->push_up();
}
int query(int l,int r){
splay(root,get_kth(root,l),null);
splay(root,get_kth(root,r+2),root);
return root->ch[1]->ch[0]->msum;
}
int find(Node *x,int k,int pos){
x->push_down();
//cout<<x->msum<<endl;
if(x->ch[0]->msum>=k)return find(x->ch[0],k,pos);
pos+=x->ch[0]->size;
if(x->val&&x->ch[0]->rsum+x->ch[1]->lsum+1>=k)return pos-x->ch[0]->rsum;
return find(x->ch[1],k,pos+1);
}
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int n,m;
while(~scanf("%d%d",&n,&m)){
init(n);
//cout<<root->msum<<" "<<root->ch[0]->msum<<" "<<root->ch[1]->msum<<endl;
while(m--){
int op,x,y;
scanf("%d",&op);
if(op==1){
scanf("%d",&x);
if(root->msum<x){
puts("0");continue;
}
int k=find(root,x,0);
printf("%d\n",k);
update(k,k+x-1,0);
}
else{
scanf("%d%d",&x,&y);
update(x,x+y-1,1);
}
}
}
return 0;
}
/*
int main()
{
int n,m;
while(cin>>n>>m) {
init(n);
while(m--){
int op,x,y,z;
cin>>op;
if(op==1){
cin>>x>>y>>z;
update(x,y,z);
}
else{
cin>>x>>y;
cout<<query(x,y)<<endl;
}
}
}
return 0;
}
*/
POJ 3667 splay区间合并练习