第k元素log(n)算法–划分树

#define M 100001
struct Seg_Tree{
    int left,right;
    int mid() {
        return (left + right) >> 1;
    }
}tt[M*4];
int len;
int sorted[M];
int toLeft[20][M];
int val[20][M];
 
void build(int l,int r,int d,int idx) {
    tt[idx].left = l;
    tt[idx].right = r;
    if(tt[idx].left == tt[idx].right)    return ;
    int mid = tt[idx].mid();
    int lsame = mid - l + 1;//lsame表示和val_mid相等且分到左边的
    for(int i = l ; i <= r ; i ++) {
        if(val[d][i] < sorted[mid]) {
            lsame --;//先假设左边的数(mid - l + 1)个都等于val_mid,然后把实际上小于val_mid的减去
        }
    }
    int lpos = l;
    int rpos = mid+1;
    int same = 0;
    for(int i = l ; i <= r ; i ++) {
        if(i == l) {
            toLeft[d][i] = 0;//toLeft[i]表示[ tt[idx].left , i ]区域里有多少个数分到左边
        } else {
            toLeft[d][i] = toLeft[d][i-1];
        }
        if(val[d][i] < sorted[mid]) {
            toLeft[d][i] ++;
            val[d+1][lpos++] = val[d][i];
        } else if(val[d][i] > sorted[mid]) {
            val[d+1][rpos++] = val[d][i];
        } else {
            if(same < lsame) {//有lsame的数是分到左边的
                same ++;
                toLeft[d][i] ++;
                val[d+1][lpos++] = val[d][i];
            } else {
                val[d+1][rpos++] = val[d][i];
            }
        }
    }
    build(l,mid,d+1,LL(idx));
    build(mid+1,r,d+1,RR(idx));
}
 
int query(int l,int r,int k,int d,int idx) {
    if(l == r) {
        return val[d][l];
    }
    int s;//s表示[ l , r ]有多少个分到左边
    int ss;//ss表示 [tt[idx].left , l-1 ]有多少个分到左边
    if(l == tt[idx].left) {
        s = toLeft[d][r];
        ss = 0;
    } else {
        s = toLeft[d][r] - toLeft[d][l-1];
        ss = toLeft[d][l-1];
    }
    if(s >= k) {//有多于k个分到左边,显然去左儿子区间找第k个
        int newl = tt[idx].left + ss;
        int newr = tt[idx].left + ss + s - 1;//计算出新的映射区间
        return query(newl,newr,k,d+1,LL(idx));
    } else {
        int mid = tt[idx].mid();
        int bb = l - tt[idx].left - ss;//bb表示 [tt[idx].left , l-1 ]有多少个分到右边
        int b = r - l + 1 - s;//b表示 [l , r]有多少个分到右边
        int newl = mid + bb + 1;
        int newr = mid + bb + b;
        return query(newl,newr,k-s,d+1,RR(idx));
    }
}
 
int main() {
    int T;
    scanf("%d",&T);
    while(T --) {
        int n , m;
        scanf("%d%d",&n,&m);
        FOR(i,1,n+1) {
            scanf("%d",&val[0][i]);
            sorted[i] = val[0][i];
        }
        sort(sorted + 1 , sorted + n + 1);
        build(1,n,0,1);
        while(m --) {
            int l,r,k;
            scanf("%d%d%d",&l,&r,&k);
            printf("%d\n",query(l,r,k,0,1));
        }
    }
    return 0;
}