bryce1010专题训练——划分树

1、求区间第K大

HDU2665 Kth number

/*划分树
查询区间第K大

*/

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;

#define ll long long
const int MAXN=100009;
int tree[21][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序好的数
int toleft[21][MAXN];//toleft[p][i]表示第i层从1到i有多少数划入左边

/*建树
same表示等于sorted[mid]的个数,初始化为mid-l+1，扫描后每出现一个更小的值，减1
所以same表示要被分入左边等于中间值的个数
建树分三种情况：
1、如果tree[[dep][i]<sorted[mid]划入左边
2、如果tree[dep][i]==sorted[mid],same>0的话划入左边
3、否则就是划入右边
*/

void build(int l,int r,int dep)
{
    if(l==r)return;
    int mid=(l+r)>>1;
    int same=mid-l+1;
    for(int i=l;i<=r;i++)
    {
        if(tree[dep][i]<sorted[mid])
        {
            same--;
        }
    }
    int lpos=l;
    int rpos=mid+1;
    for(int i=l;i<=r;i++)
    {
        if(tree[dep][i]<sorted[mid])
            tree[dep+1][lpos++]=tree[dep][i];
        else if(tree[dep][i]==sorted[mid]&&same)
        {
            tree[dep+1][lpos++]=tree[dep][i];
            same--;
        }
        else
        {
            tree[dep+1][rpos++]=tree[dep][i];
        }
        toleft[dep][i]=toleft[dep][l-1]+lpos-l;
    }
    build(l,mid,dep+1);
    build(mid+1,r,dep+1);
}
/*查询操作
查询区间第K大值，[L,R]是大区间，[l,r]是要查询的小区间
cnt记录划入左边的个数
1、如果cnt>=k，说明第K大在左子树
2、否则，说明第K大在右子树
*/
int query(int L,int R,int l,int r,int dep,int k)
{
    if(l==r)return tree[dep][l];
    int mid=(L+R)>>1;
    int cnt=toleft[dep][r]-toleft[dep][l-1];
    if(cnt>=k)
    {
        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
        int newr=newl+cnt-1;
        return query(L,mid,newl,newr,dep+1,k);
    }
    else
    {
        int newr=r+toleft[dep][R]-toleft[dep][r];
        int newl=newr-(r-l-cnt);
        return query(mid+1,R,newl,newr,dep+1,k-cnt);
    }
}

int main()
{
    int n,m;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(tree,0,sizeof(tree));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&tree[0][i]);
            sorted[i]=tree[0][i];
        }
        sort(sorted+1,sorted+n+1);
        build(1,n,0);
        int s,t,k;
        while(m--)
        {
            scanf("%d%d%d",&s,&t,&k);
            printf("%d\n",query(1,n,s,t,0,k));
        }
    }
    return 0;
}

2、查找中位数

http://acm.hdu.edu.cn/showproblem.php?pid=3473【HDU3473】

找出中间的那个x，计算公式为x*（r-l+1）/2)-(sum[l,mid]+x)+(sum[l,r]-sum[l,mid]-x)-（(r-l-前一半-1+1)/2）*x

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = 100010;
int num[20][MAXN],cnt[20][MAXN],sor[MAXN],n,leftnum;
ll sum[20][MAXN],leftsum,all[MAXN];//sum记录第d层 第i个数之前小于sor[m]的和
void build(int l,int r,int d)
{
    if(l == r){
        return ;
    }
    int m = (l + r) >> 1;
    int same_m = m - l + 1;
    for(int i = l; i <= r; i++){
        if(num[d][i] < sor[m])same_m --;
    }
    int cnt_small = 0;
    int pl,pr;
    ll val = 0;
    pl = l,pr = m + 1;
    for(int i = l; i <= r; i++){
        if(num[d][i] < sor[m]){
            cnt_small ++;
            val += num[d][i];
            sum[d][i] = val;
            cnt[d][i] = cnt_small;
            num[d+1][pl++] = num[d][i];
        }
        else if(num[d][i] == sor[m] && same_m){
            same_m --;
            cnt_small ++;
            val += num[d][i];
            sum[d][i] = val;
            cnt[d][i] = cnt_small;
            num[d+1][pl++] = num[d][i];
        }
        else {
            sum[d][i] = val;
            cnt[d][i] = cnt_small;
            num[d+1][pr++] = num[d][i];
        }
    }
    build(l,m,d+1);
    build(m+1,r,d+1);
}
ll query(int L,int R,int k,int l,int r,int d)
{
    if(l == r){
        return num[d][l];
    }
    int m = (l + r) >> 1;
    int s,ss;
    ll val = 0;
    if(l == L)s = 0, val = sum[d][R];
    else s = cnt[d][L-1], val = sum[d][R] - sum[d][L-1];
    ss = cnt[d][R] - s;
    if(ss >= k){
        int newl = l + s;
        int newr = l + s + ss - 1;
        return query(newl,newr,k,l,m,d+1);
    }
    else {
        leftnum += ss;//进入左孩子不用处理，进入右孩子时 就要加上左孩子的值
        leftsum += val;
        int a = L - l - s;
        int b = R - L + 1 - ss;
        int newl = m + 1 + a;
        int newr = m + 1 + a + b - 1;
        return query(newl,newr,k - ss,m+1,r,d+1);
    }
}
int main()
{
    int t,ff = 0;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        memset(all,0,sizeof(all));
        memset(num,0,sizeof(num));
        for(int i = 1; i <= n; i++){
            scanf("%d",&num[1][i]);
            sor[i] = num[1][i];
            all[i] = all[i-1] + sor[i];
        }
        sort(sor + 1,sor + n + 1);
        build(1,n,1);
        int q,x,y;
        scanf("%d",&q);
        printf("Case #%d:\n",++ff);
        while(q--){
            scanf("%d%d",&x,&y);
            x += 1;
            y += 1;
            int len = (y - x + 1);
            ll tp;
            leftnum = 0;
            leftsum = 0;
            if(len % 2){
                int k = (len + 1) >> 1;
                tp = query(x,y,k,1,n,1);
            }
            else {
                int k = len >> 1;
                tp = query(x,y,k,1,n,1);
            }
            //cout<<tp<<‘ ‘<<leftnum<<‘ ‘<<leftsum<<‘ ‘<<all[y]<<‘ ‘<<all[x+leftnum]<<endl;
            ll ans = tp * (leftnum + 1) - (leftsum + tp) + (all[y] - all[x - 1] - (leftsum + tp)) - (y - x + 1 - (leftnum + 1)) * tp;
            printf("%lld\n",ans);
        }
        printf("\n");
    }
    return 0;
}