hdu4251The Famous ICPC Team Again【划分树入门题】

水到不能再水……

/***************hdu42512016.1.281060MS 15816K 2320 B G++***************/#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 1000010;int sorted[MAXN];   //对原集合中元素排序后的值int val[30][MAXN];  //val记录第k层当前位置的值int toleft[30][MAXN];  //记录元素所在区间当前位置前的元素进入到左子树的个数int sum[30][MAXN];  //记录比当前元素小的元素的和int n;void build(int l, int r, int d) {    if (l == r) return ;    int mid = (l + r) >> 1;    int same = mid - l + 1;    for (int i=l; i<=r; i++)        if (val[d][i] < sorted[mid])            same--;    int lp = l, rp = mid+1;    for (int i=l; i<=r; i++) {        if (i == l) toleft[d][i] = 0;        else toleft[d][i] = toleft[d][i-1];        if (val[d][i] < sorted[mid]) {            toleft[d][i]++;            val[d+1][lp++] = val[d][i];        } else if (val[d][i] > sorted[mid])            val[d+1][rp++] = val[d][i];        else {            if (same) {                same--;                toleft[d][i]++;                val[d+1][lp++] = val[d][i];            } else val[d+1][rp++] = val[d][i];        }    }    build(l, mid, d+1);    build(mid+1, r, d+1);}int query(int a, int b, int k, int l, int r, int d) {    if (a == b) return val[d][a];    int mid = (l + r) >> 1;    int s, ss, sss;    if (a == l) {        s = toleft[d][b];        ss = 0;    } else {        s = toleft[d][b] - toleft[d][a-1];        ss = toleft[d][a-1];    }    if (s >= k) {        a = l + ss;        b = l + ss + s - 1;        return query(a, b, k, l, mid, d+1);    } else {        a = mid+1 + a - l - ss;        b = mid+1 + b - l - toleft[d][b];        return query(a, b, k-s, mid+1, r, d+1);    }}int main(){    //freopen("cin.txt","r",stdin);     int n, m,tot=1;    while(~scanf("%d", &n))    {        printf("Case %d:\n",tot++);        for (int i=1; i<=n; i++) {            scanf("%d", &sorted[i]);            val[0][i] = sorted[i];        }        sort(sorted+1, sorted+1+n);        build(1, n, 0);        int a, b, k;        scanf("%d",&m);        while (m--) {            scanf("%d%d", &a, &b);            printf("%d\n", query(a, b, (b-a)/2+1, 1, n, 0));        }    }    return 0;}