允许重复数字的划分树

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const int MAXN = 50010,MAXM = 20;

int num[MAXN],tmp[MAXN],pre[MAXN];

int ans[MAXM][MAXN],val[MAXM][MAXN];

void Init(int l,int r,int h)
{
    if(l == r)
        return ;

    int mid = (l+r)>>1;

    int i,pl,pr,sum;

    for(i = l,pl = l,pr = mid+1,sum = 0;i <= r; ++i)
    {
        if(val[h][i] < tmp[mid] || (val[h][i] == tmp[mid] && sum < min(pre[mid],mid-l+1)))
        {
            val[h+1][pl++] = val[h][i],ans[h][i] = (i == l ? 1 : ans[h][i-1]+1) ;
            if(val[h][i] == tmp[mid])
                sum++;
        }
        else
            val[h+1][pr++] = val[h][i],ans[h][i] = (i == l ? 0 : ans[h][i-1]);
    }

    Init(l,mid,h+1);
    Init(mid+1,r,h+1);
}

int Query(int L,int R,int l,int r,int k,int h)
{
    if(L == R)
        return val[h][L];

    int goleftr = ans[h][r] ;//[L,r]内被分到左边的
    int goleftl = (l == L ? 0 : ans[h][l-1]);//[L,l-1]内被分到左边的
    int goleft = goleftr-goleftl;//[l,r]内被分到左边的
    int mid = (L+R)>>1;
    int gorightr = r-L+1 - ans[h][r];//[L,r]内被分到右边的
    int gorightl = l-L-goleftl;//[L,l-1]内被分到右边的

    if(k <= goleft)
        return Query(L,mid,L-1+goleftl+1,L-1+goleftr,k,h+1);
    return Query(mid+1,R,mid+gorightl+1,mid+gorightr,k-goleft,h+1);
}

void InitKthTree(int n)
{
        memcpy(tmp,num,sizeof(num));
        memcpy(val[0],num,sizeof(num));

        sort(tmp+1,tmp+n+1);
        int i;
        for(i = 2,pre[1] = 1;i <= n; ++i)
            pre[i] = (tmp[i] == tmp[i-1]  ?  pre[i-1]+1 : 1);

        Init(1,n,0);
}

允许重复数字的划分树

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原文地址：http://blog.csdn.net/zmx354/article/details/44943163