HDU4973:A simple simulation problem.(线段树)

Problem Description

There are n types of cells in the lab, numbered from 1 to n. These cells are put in a queue, the i-th cell belongs to type i. Each time I can use mitogen to double the cells in the interval [l, r]. For instance, the original queue is {1 2 3 3 4 5}, after using a mitogen in the interval [2, 5] the queue will be {1 2 2 3 3 3 3 4 4 5}. After some operations this queue could become very long, and I can’t figure out maximum count of cells of same type. Could you help me?

 

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases.

For each case, the first line contains 2 integers (1 <= n,m<= 50000) indicating the number of cell types and the number of operations.

For the following m lines, each line represents an operation. There are only two kinds of operations: Q and D. And the format is:

“Q l r”, query the maximum number of cells of same type in the interval [l, r];
“D l r”, double the cells in the interval [l, r];

(0 <= r – l <= 10^8, 1 <= l, r <= the number of all the cells)

 

Output

For each case, output the case number as shown. Then for each query "Q l r", print the maximum number of cells of same type in the interval [l, r].

Take the sample output for more details.

 

Sample Input

15 5D 5 5Q 5 6D 2 3D 1 2Q 1 7

 

Sample Output

Case #1:23
#include<stdio.h>#define L 50005#define ll __int64#define ls 2*i#define rs 2*i+1#define w(x) while(x)#define max(a,b) (a>b?a:b)int t,n,m,cas=1;ll x,y,ans,res;char str[10];struct node{    int l,r,flag;    ll maxn,sum;} a[L<<2];void PushUp(int i){    a[i].sum=a[2*i].sum+a[2*i+1].sum;    a[i].maxn=max(a[2*i].maxn,a[2*i+1].maxn);}void PushDown(int i){    if(a[i].flag)    {        a[ls].flag += a[i].flag;        a[rs].flag += a[i].flag;        w(a[i].flag--)        {            a[ls].maxn *= 2;            a[rs].maxn *= 2;            a[ls].sum *= 2;            a[rs].sum *= 2;        }        a[i].flag=0;    }}void init(int l,int r,int i){    a[i].l=l;    a[i].r=r;    a[i].flag=0;    a[i].maxn=1;    if(l==r)    {        a[i].sum=1;        return;    }    int mid=(l+r)/2;    init(l,mid,ls);    init(mid+1,r,rs);    PushUp(i);}int find(int i,ll x,ll &k,ll &tem){    if(a[i].l==a[i].r)    {        k=x;        tem=a[i].sum;        return a[i].l;    }    PushDown(i);    if(x<=a[ls].sum) return find(ls,x,k,tem);    else return find(rs,x-a[ls].sum,k,tem);}void insert1(int i,int p,ll v){    if(a[i].l==a[i].r)    {        a[i].maxn+=v;        a[i].sum+=v;        return;    }    PushDown(i);    int mid=(a[i].l+a[i].r)/2;    if(p<=mid) insert1(ls,p,v);    else insert1(rs,p,v);    PushUp(i);}void insert2(int l,int r,int i){    if(l<=a[i].l&&a[i].r<=r)    {        PushDown(i);        a[i].flag++;        a[i].sum*=2;        a[i].maxn*=2;        return;    }    PushDown(i);    int mid=(a[i].l+a[i].r)/2;    if(l<=mid) insert2(l,r,ls);    if(r>mid) insert2(l,r,rs);    PushUp(i);}void query(int l,int r,int i){    if(l<=a[i].l&&a[i].r<=r)    {        res=max(res,a[i].maxn);        return;    }    PushDown(i);    int mid=(a[i].l+a[i].r)/2;    if(l<=mid) query(l,r,ls);    if(r>mid) query(l,r,rs);}int main(){    scanf("%d",&t);    w(t--)    {        scanf("%d%d",&n,&m);        init(1,n,1);        printf("Case #%d:\n",cas++);        w(m--)        {            scanf("%s%I64d%I64d",str,&x,&y);            ll lk,rk,tem;            int r=find(1,y,rk,tem);            int l=find(1,x,lk,tem);            if(str[0]=='D')            {                if(l==r)                {                    insert1(1,l,rk-lk+1);                    continue;                }                insert1(1,r,rk);                insert1(1,l,tem-lk+1);                if(r-l>1)                    insert2(l+1,r-1,1);            }            else            {                ans=0;                if(l==r)                {                    printf("%I64d\n",rk-lk+1);                    continue;                }                ans=max(ans,max(rk,tem-lk+1));                if(r-l>1)                {                    res=0;                    query(l+1,r-1,1);                    ans=max(ans,res);                }                printf("%I64d\n",ans);            }        }    }    return 0;}