HDU 2871 Memory Control (線段樹，區間合并)，hdu2871

HDU 2871 Memory Control (線段樹，區間合并)，hdu2871

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

Memory Control Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4418    Accepted Submission(s): 1056


Problem DescriptionMemory 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. 
 
InputInput 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.
 
OutputFor 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 10New 2New 5New 2New 2Free 3Get 1Get 2Get 3Free 3Reset

 
Sample Output

New at 1Reject NewNew at 3New at 5 Free from 3 to 4Get at 1Get at 5Reject GetReject FreeReset Now

 
Source2009 Multi-University Training Contest 7 - Host by FZU 

題意：

Reset:清空記憶體

New x:申請長度為x的記憶體

Free x:釋放x所在的記憶體塊

Get x:詢問x所在記憶體塊的起點

分析：

New操作用線段樹，找到能容納x的最左的位置，維護節點的左/右起最長連續0的長度和對應區間內最長連續0的長度。

把現有記憶體塊儲存，用二分尋找/更新即可，注意Reset時要清空。

#include<cstdio>#include<iostream>#include<cstdlib>#include<algorithm>#include<ctime>#include<cctype>#include<cmath>#include<string>#include<cstring>#include<stack>#include<queue>#include<list>#include<vector>#include<map>#include<set>#define sqr(x) ((x)*(x))#define LL long long#define itn int#define INF 0x3f3f3f3f#define PI 3.1415926535897932384626#define eps 1e-10#define maxm#define maxn 200007using namespace std;int setv[maxn<<2];int lsum0[maxn<<2],rsum0[maxn<<2],msum0[maxn<<2];struct Memory{    int l,r;};vector<Memory> v;inline void pushup(int k,int l,int r){    int lc=k*2+1,rc=k*2+2,m=l+r>>1;    if (lsum0[lc]==m-l)   lsum0[k]=lsum0[lc]+lsum0[rc]; else    lsum0[k]=lsum0[lc];    if (rsum0[rc]==r-m)   rsum0[k]=rsum0[rc]+rsum0[lc]; else    rsum0[k]=rsum0[rc];    msum0[k]=max(rsum0[lc]+lsum0[rc],max(msum0[lc],msum0[rc]));}inline void pushdown(int k,int l,int r){    if (setv[k]!=-1)    {        int lc=k*2+1,rc=k*2+2,m=l+r>>1;        setv[lc]=setv[rc]=setv[k];        lsum0[lc]=rsum0[lc]=msum0[lc]=setv[k]?0:m-l;        lsum0[rc]=rsum0[rc]=msum0[rc]=setv[k]?0:r-m;        setv[k]=-1;    }}void update(int a,int b,int v,int k,int l,int r){    if (b<=l || r<=a)   return ;    if (a<=l && r<=b)    {        setv[k]=v;        lsum0[k]=rsum0[k]=msum0[k]=v?0:r-l;    }    else    {        pushdown(k,l,r);        update(a,b,v,k*2+1,l,l+r>>1);        update(a,b,v,k*2+2,l+r>>1,r);        pushup(k,l,r);    }}int query(int w,int k,int l,int r){    int lc=k*2+1,rc=k*2+2,m=l+r>>1;    if (r-l==1) return l;    if (r-l!=1) pushdown(k,l,r);    if (msum0[lc]>=w) return query(w,lc,l,m);    if (rsum0[lc]+lsum0[rc]>=w)     return m-rsum0[lc];    return query(w,rc,m,r);}int bin_search(int k){    int l=0,r=v.size()-1,ans=-1;    while (l<=r)    {        itn mid=l+r>>1;        if (v[mid].l<=k)        {            l=mid+1;            ans=mid;        }        else        {            r=mid-1;        }    }    return ans;}int main(){    #ifndef ONLINE_JUDGE        freopen("/home/fcbruce/文檔/code/t","r",stdin);    #endif // ONLINE_JUDGE    int n,m,w,a,b,p;    char op[10];    while (~scanf("%d %d",&n,&m))    {        v.clear();        update(0,n,0,0,0,n);        while (m--)        {            scanf("%s",op);            if (op[0]=='R')            {                v.clear();                puts("Reset Now");                update(0,n,0,0,0,n);                continue;            }            if (op[0]=='N')            {                scanf("%d",&w);                if (w<=msum0[0])                {                    a=query(w,0,0,n);                    b=a+w;                    update(a,b,1,0,0,n);                    printf("New at %d\n",a+1);                    Memory temp;                    temp.l=a+1;                    temp.r=b;                    int id=bin_search(a+1)+1;                    v.insert(v.begin()+id,temp);                }                else                    puts("Reject New");                continue;            }            if (op[0]=='F')            {                scanf("%d",&p);                int id=bin_search(p);                if (id==-1 || v[id].r<p)                    puts("Reject Free");                else                {                    printf("Free from %d to %d\n",v[id].l,v[id].r);                    update(v[id].l-1,v[id].r,0,0,0,n);                    v.erase(v.begin()+id,v.begin()+id+1);                }                continue;            }            if (op[0]=='G')            {                scanf("%d",&w);                if (w<=v.size())                    printf("Get at %d\n",v[w-1].l);                else                    puts("Reject Get");            }        }        puts("");    }    return 0;}