hdu 2871 Memory Control(线段树的各种操作&Splay解法)

Memory Control

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2043    Accepted Submission(s): 480

Problem Description

Memory units are numbered from 1 up to N.
A sequence of memory units is called a memory block. 
The memory control system we consider now has four kinds of operations:
1.  Reset Reset all memory units free.
2.  New x Allocate a memory block consisted of x continuous free memory units with the least start number
3.  Free x Release the memory block which includes unit x
4.  Get x Return the start number of the xth memory block(Note that we count the memory blocks allocated from left to right)
Where 1<=x<=N.You are request to find out the output for M operations. 

 

Input

Input contains multiple cases.
Each test case starts with two integer N,M(1<=N,M<=50000) ,indicating that there are N units of memory and M operations.
Follow by M lines,each line contains one operation as describe above.

 

Output

For each “Reset” operation, output “Reset Now”.
For each “New” operation, if it’s possible to allocate a memory block,
output “New at A”,where Ais the least start number,otherwise output “Reject New”.
For each “Free” operation, if it’s possible to find a memory block occupy unit x,
output “Free from A to B”,where A and B refer to the start and end number of the memory block,otherwise output “Reject Free”.
For each “Get” operation, if it’s possible to find the xth memory blocks,
output “Get at A”,where A is its start number,otherwise output “Reject Get”.
Output one blank line after each test case.

 

Sample Input

   
   
   
   

    
    
    
    6 10
New 2
New 5
New 2
New 2
Free 3
Get 1
Get 2
Get 3
Free 3
Reset
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    New at 1
Reject New
New at 3
New at 5 
Free from 3 to 4
Get at 1
Get at 5
Reject Get
Reject Free
Reset Now
   
   
   
   

 

Source

2009 Multi-University Training Contest 7 - Host by FZU

 

Recommend

gaojie

 

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2871

题意：让你模拟内存控制，可能申请连续x个空间，返回编号最小的位置，或者清空包含x单元的整个空间，返回空间的头和尾，或者获取从左到右数的第x的空间，或者整个内存重置为空。。。。

分析：申请空间和释放空间，和这题很像，只要申请的时候把被覆盖改成被哪个id覆盖，然后开个数组记录这个id的头和尾，释放的时候就可以找到覆盖的id，进而找到头和尾，重置内存其实就是把整个内存都释放掉。。。不用多写函数。。。至于寻找第x个空间，这个比较麻烦，如果另开一棵线段树来统计，我想回好写很多，就单点更新和成段清零，用个延迟标记就行，不过我最后选择直接写在一起了，这个就得有两个延迟标记，一个记录是否要累加，一个记录是否清零，当然，清零的时候累加标记也要清零。。。

一开始少了个清零标记，把清零和累加混在一起了，wa了无数次啊。。。反正写起来实在是麻烦了，或许写成两棵线段树就好很多了，一开始的想法就是两棵线段树的，早上突然想写成一棵 = =，不过别人貌似都是用vector记录。。。。表示很少用那东西- -

2012-10-7：

这题用splay来搞，简直是自虐啊，我调试了整整一个晚上。。。

其实思路就是照搬线段树的那些标记，剩下的就是splay的更新操作边界处理比较麻烦，一般都加上最大和最小两个节点来忽略边界，不过这题这样做似乎导致更多的麻烦T_T

反正我是调到死了。。。400++ms过的，代码巨挫，不想多说了，累了，具体看看代码吧，很容易懂的

代码：

#include<cstdio>
#include<iostream>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define uprt rt,m-l+1,r-m
using namespace std;
const int mm=55555;
const int mn=mm<<2;
int dly[mn],add[mn],clr[mn],sum[mn],ml[mn],lm[mn],rm[mn];
int sl[mm],sr[mm];
void set(int rt,int id,int len)
{
    ml[rt]=lm[rt]=rm[rt]=id?0:len;
    if(!id)sum[rt]=add[rt]=0,clr[rt]=1;
    dly[rt]=id;
}
void pushdown(int rt,int l1,int l2)
{
    if(dly[rt]>-1)
    {
        set(rt<<1,dly[rt],l1);
        set(rt<<1|1,dly[rt],l2);
        dly[rt]=-1;
    }
    if(clr[rt])
    {
        sum[rt<<1]=sum[rt<<1|1]=add[rt<<1]=add[rt<<1|1]=0;
        clr[rt<<1]=clr[rt<<1|1]=1;
        clr[rt]=0;
    }
    if(add[rt])
    {
        ++sum[rt<<1],add[rt<<1]=1;
        add[rt]=0;
    }
}
void pushup(int rt,int l1,int l2)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
    ml[rt]=max(rm[rt<<1]+lm[rt<<1|1],max(ml[rt<<1],ml[rt<<1|1]));
    lm[rt]=lm[rt<<1],rm[rt]=rm[rt<<1|1];
    if(lm[rt]>=l1)lm[rt]+=lm[rt<<1|1];
    if(rm[rt]>=l2)rm[rt]+=rm[rt<<1];
}
void build(int l,int r,int rt)
{
    dly[rt]=-1,clr[rt]=add[rt]=sum[rt]=0;
    ml[rt]=lm[rt]=rm[rt]=r-l+1;
    if(l==r)return;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void updata(int L,int R,int id,int l,int r,int rt)
{
    if(L<=l&&R>=r)
    {
        set(rt,id,r-l+1);
        if(id&&L==l)++sum[rt],add[rt]=1;
        return;
    }
    int m=(l+r)>>1;
    pushdown(uprt);
    if(L<=m)updata(L,R,id,lson);
    if(R>m)updata(L,R,id,rson);
    pushup(uprt);
}
int NEW(int x,int l,int r,int rt)
{
    if(lm[rt]>=x)return l;
    int m=(l+r)>>1,ret;
    pushdown(uprt);
    if(ml[rt<<1]>=x)ret=NEW(x,lson);
    else if(rm[rt<<1]+lm[rt<<1|1]>=x)ret=m-rm[rt<<1]+1;
    else ret=NEW(x,rson);
    pushup(uprt);
    return ret;
}
int FREE(int x,int l,int r,int rt)
{
    if(dly[rt]>-1)return dly[rt];
    if(l==r)return 0;
    int m=(l+r)>>1,ret;
    pushdown(uprt);
    if(x<=m)ret=FREE(x,lson);
    else ret=FREE(x,rson);
    pushup(uprt);
    return ret;
}
int GET(int x,int l,int r,int rt)
{
    if(l==r)return l;
    int m=(l+r)>>1,ret;
    pushdown(uprt);
    if(sum[rt<<1]>=x)ret=GET(x,lson);
    else ret=GET(x-sum[rt<<1],rson);
    pushup(uprt);
    return ret;
}
int main()
{
    int x,n,m,top;
    char op[55];
    while(~scanf("%d%d",&n,&m))
    {
        build(1,n,1);
        top=0;
        while(m--)
        {
            scanf("%s",op);
            if(op[0]=='R')
            {
                updata(1,n,0,1,n,1);
                puts("Reset Now");
            }
            else scanf("%d",&x);
            if(op[0]=='N')
            {
                if(ml[1]>=x)
                {
                    ++top;
                    sl[top]=NEW(x,1,n,1);
                    sr[top]=sl[top]+x-1;
                    updata(sl[top],sr[top],top,1,n,1);
                    printf("New at %d\n",sl[top]);
                }
                else puts("Reject New");
            }
            if(op[0]=='F')
            {
                x=FREE(x,1,n,1);
                if(x)
                {
                    updata(sl[x],sr[x],0,1,n,1);
                    printf("Free from %d to %d\n",sl[x],sr[x]);
                }
                else puts("Reject Free");
            }
            if(op[0]=='G')
            {
                if(sum[1]>=x)printf("Get at %d\n",GET(x,1,n,1));
                else puts("Reject Get");
            }
        }
        puts("");
    }
    return 0;
}

splay代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int mm=55555;
int l[mm],r[mm];
struct SplayTree
{
    int son[mm][2],far[mm],num[mm],val[mm],len[mm],blk[mm],ll[mm],rl[mm],lv[mm],rv[mm],dly[mm];
    int rt,size,col;
    void Link(int x,int y,int c)
    {
        far[x]=y,son[y][c]=x;
    }
    void Rotate(int x,int c)
    {
        int y=far[x];
        PushDown(y);
        PushDown(x);
        Link(x,far[y],son[far[y]][1]==y);
        Link(son[x][!c],y,c);
        Link(y,x,!c);
        PushUp(y);
    }
    void Splay(int x,int g)
    {
        for(PushDown(x);far[x]!=g;)
        {
            int y=far[x],cx=son[y][1]==x,cy=son[far[y]][1]==y;
            if(far[y]==g)Rotate(x,cx);
            else
            {
                if(cx==cy)Rotate(y,cy);
                else Rotate(x,cx);
                Rotate(x,cy);
            }
        }
        PushUp(x);
        if(!g)rt=x;
    }
    int Select(int k,int g)
    {
        int x=rt;
        PushDown(x);
        while(num[son[x][0]]!=k)
        {
            if(num[son[x][0]]>k)x=son[x][0];
            else k-=num[son[x][0]]+1,x=son[x][1];
            PushDown(x);
        }
        Splay(x,g);
        return x;
    }
    void NewNode(int y,int &x)
    {
        x=++size;
        far[x]=y,num[x]=1,ll[x]=rl[x]=len[x]=1;
        blk[x]=lv[x]=rv[x]=val[x]=dly[x]=son[x][0]=son[x][1]=0;
    }
    void Make(int l,int r,int &x,int y)
    {
        if(l>r)return;
        int m=(l+r)>>1;
        NewNode(y,x);
        Make(l,m-1,son[x][0],x);
        Make(m+1,r,son[x][1],x);
        PushUp(x);
    }
    void Prepare(int n)
    {
        NewNode(size=0,rt);
        NewNode(rt,son[rt][1]);
        val[1]=val[2]=-1,len[1]=len[2]=0;
        lv[1]=rv[1]=lv[2]=rv[2]=0;
        col=ll[1]=rl[1]=ll[2]=rl[2]=0;
        Make(1,n,son[2][0],2);
        Splay(2,0);
    }
    void Check(int x,int d)
    {
        if(!x)return;
        dly[x]=d;
        len[x]=ll[x]=rl[x]=(d<0)*num[x];
        val[x]=lv[x]=rv[x]=d>0?d:0;
        blk[x]=d>0;
    }
    void PushDown(int x)
    {
        if(!dly[x])return;
        Check(son[x][0],dly[x]);
        Check(son[x][1],dly[x]);
        dly[x]=0;
    }
    void PushUp(int x)
    {
        int a=son[x][0],b=son[x][1];
        num[x]=1+num[a]+num[b];
        len[x]=max(len[a],len[b]);
        blk[x]=blk[a]+blk[b];
        if(rv[a]&&rv[a]==lv[b])--blk[x];
        ll[x]=ll[a],rl[x]=rl[b];
        lv[x]=a>2?lv[a]:val[x],rv[x]=b>2?rv[b]:val[x];
        if(!val[x])
        {
            len[x]=max(len[x],rl[a]+ll[b]+1);
            if(len[a]>=num[a])ll[x]+=ll[b]+1;
            if(len[b]>=num[b])rl[x]+=rl[a]+1;
        }
        else if(x>2&&val[x]!=rv[a]&&val[x]!=lv[b])++blk[x];
    }
    void NEW(int a)
    {
        if(len[rt]<a)
        {
            puts("Reject New");
            return;
        }
        int x=rt,y,b=a,p=0;
        PushDown(x);
        while(x)
        {
            if(len[son[x][0]]>=b)x=son[x][0];
            else if(rl[son[x][0]]+!val[x]==b)break;
            else
            {
                if(!val[x]&&rl[son[x][0]]+ll[son[x][1]]+1>=b)b-=rl[son[x][0]]+1;
                p+=num[son[x][0]]+1,x=son[x][1];
            }
            PushDown(x);
        }
        p+=num[son[x][0]]-a+1;
        x=Select(p-1,0);
        y=Select(p+a,rt);
        Check(son[y][0],++col);
        Splay(y,0);
        l[col]=p,r[col]=p+a-1;
        printf("New at %d\n",p);
    }
    void FREE(int x)
    {
        x=Select(x,0);
        if(val[x])
        {
            int t=val[x];
            Select(l[t]-1,0);
            int y=Select(r[t]+1,rt);
            Check(son[y][0],-1);
            Splay(y,0);
            printf("Free from %d to %d\n",l[t],r[t]);
        }
        else puts("Reject Free");
    }
    void GET(int a)
    {
        if(a>blk[rt])
        {
            puts("Reject Get");
            return;
        }
        int x=rt,y;
        PushDown(x);
        while(x)
        {
            y=son[x][0];
            if(blk[y]>=a)x=y;
            else if(x>2&&val[x]&&val[x]!=rv[y]&&blk[y]+1==a)break;
            else
            {
                a-=blk[y];
                if(x>2&&val[x])
                {
                    if(val[x]!=rv[y]&&val[x]!=lv[son[x][1]])--a;
                    if(val[x]==rv[y]&&val[x]==lv[son[x][1]])++a;
                }
                x=son[x][1];
            }
            PushDown(x);
        }
        Splay(x,0);
        printf("Get at %d\n",l[val[x]]);
    }
    void Reset()
    {
        Splay(1,0);
        Splay(2,1);
        Check(son[2][0],-1);
        Splay(2,0);
        puts("Reset Now");
    }
}spt;
int i,n,m;
char op[22];
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        spt.Prepare(n);
        while(m--)
        {
            scanf("%s",op);
            if(op[0]!='R')scanf("%d",&i);
            if(op[0]=='N')spt.NEW(i);
            if(op[0]=='F')spt.FREE(i);
            if(op[0]=='G')spt.GET(i);
            if(op[0]=='R')spt.Reset();
        }
        puts("");
    }
    return 0;
}