hdu5372 离散化+树状数组

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这个题离散化有点蛋疼。。。
代码：
#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#include<string>#include<iomanip>#pragma comment(linker, "/STACK:1024000000,1024000000")#include<vector>#include<set>#include<map>#include<queue>#include<list>using namespace std;typedef long long LL;typedef unsigned long long ULL;#define rep(i,k,n) for(int i=(k);i<=(n);i++)#define red(i,k,n) for(int i=(k);i>=(n);i--)#define sqr(x) ((x)*(x))#define clr(x,y) memset((x),(y),sizeof(x))#define mod 1000000007#define MAX(a,b) ((a)>(b)?(a):(b))#define MIN(a,b) ((a)>(b)?(b):(a))const int maxn = 400010;struct opp{    bool type;    int de;} op[200010];int a[maxn],n,N,acnt;int r[maxn];bool cmp(int x,int y){    return a[x]<a[y];}void dis(){    rep(i,1,acnt)r[i]=i;    sort(r+1,r+1+acnt,cmp);    int pre=a[r[1]];    N=a[r[1]]=1;    rep(i,2,acnt)    {        if(a[r[i]]==pre)a[r[i]]=a[r[i-1]];        else        {            pre=a[r[i]];            N=a[r[i]]=a[r[i-1]]+1;        }    }}struct Bit{    int bit[maxn];    void cl()    {        clr(bit,0);    }    void add(int k,int v)    {        for(int i=k; i<=N; i+=i&-i)bit[i]+=v;    }    int sum(int k)    {        int ret=0;        for(int i=k; i>0; i-=i&-i)ret+=bit[i];        return ret;    }} BL,BR;int main(){//#define LOCAL#ifdef LOCAL    freopen("e:\\read.txt","r",stdin);    //freopen("e:\\write.txt","w",stdout);#endif    int ii=0;    while(~scanf("%d",&n))    {        acnt=0;        rep(i,1,n)        {            int t;            scanf("%d",&t);            op[i].type=t;            if(t)            {                scanf("%d",&op[i].de);            }            else            {                acnt+=2;                scanf("%d",&t);                a[acnt-1]=t;                a[acnt]=t+acnt/2;            }        }        dis();        BL.cl();        BR.cl();        int s=1;        printf("Case #%d:\n",++ii);        rep(i,1,n)        {            if(op[i].type)            {                int k=op[i].de;                BL.add(a[k*2-1],-1);                BR.add(a[k*2],-1);            }            else            {                printf("%d\n",BR.sum(a[2*s])-BL.sum(a[2*s-1]-1));                BL.add(a[2*s-1],1);                BR.add(a[2*s],1);                s++;            }        }    }    return 0;}
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