HDU ACM 5249 kpi-> tree-shaped array + two-point

Analysis: Note that the queue holds the subscript in the sorted array. The tree array is also processed according to the subscript.

#include <iostream> #include <queue> #include <algorithm>using namespace std; #define N 10005int C[n <<2];int N,a[n],b[n];char str[n][10];int lowbit (int x) {return x& (-X);}        void Update (int x,int val) {while (x<n) {c[x]+=val;    X+=lowbit (x);    }}int sum (int x) {int s=0;        while (x>0) {s+=c[x];    X-=lowbit (x); } return s;}    int solve (int x) {int id,l,r,mid;    id=x/2+1;    l=1;    R=n;        while (l<=r) {mid= (l+r) >>1;        if (sum (mid) >=id) r=mid-1;    else l=mid+1; } return b[l];}    int main () {int x,t=0,i,cnt;        while (scanf ("%d", &n) ==1) {queue<int> q;        Memset (C,0,sizeof (c));        Cnt=1;            for (i=1;i<=n;i++) {scanf ("%s", Str[i]);                if (str[i][0]== ' I ') {scanf ("%d", &x);                B[cnt++]=x;            A[i]=x; }} sort (b+1,b+cnt+1);         Sort Cnt=unique (b+1,b+cnt+1)-(b+1);        Go to the heavy printf ("Case #%d:\n", ++t);                 for (i=1;i<=n;i++) {if (str[i][0]== ' I ') {X=lower_bound (b+1,b+cnt+1,a[i])-B;                Update (x,1);           Q.push (x); The queue holds the subscript} else if (str[i][0]== ' O ') {x=lower_bound (B+1,b+cnt+1,b[q.fron                T ()])-B;                Update (X,-1);            Q.pop ();        } else if (str[i][0]== ' Q ') printf ("%d\n", Solve (Q.size ()));      }} return 0; }

HDU ACM 5249 kpi-> tree-shaped array + two-point