[BZOJ 1901] Dynamic Rankings 【树状数组套线段树 || 线段树套线段树】

题目链接：BZOJ - 1901

 

题目分析

树状数组套线段树或线段树套线段树都可以解决这道题。

第一层是区间，第二层是权值。

空间复杂度和时间复杂度均为 O(n log^2 n)。

线段树比树状数组麻烦好多...我容易写错= =

 

代码

树状数组套线段树

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#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
 
using namespace std;
 
const int MaxN = 10000 + 5, MN = 1000000015, MaxNode = 10000 * 30 * 15 + 15;
 
int n, m, Index, Used_Index;
int A[MaxN], Root[MaxN], Son[MaxNode][2], T[MaxNode], U[MaxN], C[MaxN];
 
void Add(int &x, int s, int t, int Pos, int Num)
{
    if (x == 0) x = ++Index;
    T[x] += Num;
    if (s == t) return;
    int m = (s + t) >> 1;
    if (Pos <= m) Add(Son[x][0], s, m, Pos, Num);
    else Add(Son[x][1], m + 1, t, Pos, Num);
}
 
void Change(int x, int Pos, int Num)
{
    for (int i = x; i <= n; i += i & -i)
        Add(Root[i], 0, MN, Pos, Num);
}
 
int Get_Sum(int x)
{
    int ret = 0;
    for (int i = x; i; i -= i & -i)
        ret += T[Son[U[i]][0]];
    return ret;
}
 
void Init_U(int x)
{
    for (int i = x; i; i -= i & -i)
        U[i] = Root[i];
}
 
void Turn(int x, int f)
{
    for (int i = x; i; i -= i & -i)
    {
        if (C[i] == Used_Index) break;
        C[i] = Used_Index;
        U[i] = Son[U[i]][f];
    }
}
 
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &A[i]);
        Change(i, A[i], 1);
    }
    char f;
    int Pos, Num, L, R, k, Temp;
    for (int i = 1; i <= m; ++i)
    {
        f = '-';
        while (f < 'A' || f > 'Z') f = getchar();
        if (f == 'C')
        {
            scanf("%d%d", &Pos, &Num);
            Change(Pos, A[Pos], -1);
            A[Pos] = Num;
            Change(Pos, Num, 1);
        }
        else
        {
            scanf("%d%d%d", &L, &R, &k);
            int l, r, mid;
            l = 0; r = MN;
            Init_U(L - 1);
            Init_U(R);
            Used_Index = 0;
            while (l < r)
            {
                mid = (l + r) >> 1;
                Temp = Get_Sum(R) - Get_Sum(L - 1);
                ++Used_Index;
                if (Temp >= k)
                {
                    r = mid;
                    Turn(L - 1, 0);
                    Turn(R, 0);
                }
                else
                {
                    l = mid + 1;
                    k -= Temp;
                    Turn(L - 1, 1);
                    Turn(R, 1);
                }
            }
            printf("%d\n", l);
        }
    }
    return 0;
}

 

线段树套线段树

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#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
 
using namespace std;
 
const int MaxN = 10000 + 5, MN = 1000000000 + 15, MaxNode = 10000 * 30 * 15 + 15;
 
int n, m, Index, Used_Index;
int A[MaxN], Root[MaxN * 4], T[MaxNode], Son[MaxNode][2], U[MaxN * 4], C[MaxN * 4];
 
void Add(int &x, int s, int t, int Pos, int Num)
{
    if (x == 0) x = ++Index;
    T[x] += Num;
    if (s == t) return;
    int m = (s + t) >> 1;
    if (Pos <= m) Add(Son[x][0], s, m, Pos, Num);
    else Add(Son[x][1], m + 1, t, Pos, Num);
}
 
void Change(int x, int s, int t, int Pos, int Pos_2, int Num)
{
    Add(Root[x], 0, MN, Pos_2, Num);
    if (s == t) return;
    int m = (s + t) >> 1;
    if (Pos <= m) Change(x << 1, s, m, Pos, Pos_2, Num);
    else Change(x << 1 | 1, m + 1, t, Pos, Pos_2, Num);
}
 
void Init_U(int x, int s, int t, int Pos)
{
    if (Pos >= t)
    {
        U[x] = Root[x];
        return;
    }
    int m = (s + t) >> 1;
    Init_U(x << 1, s, m, Pos);
    if (Pos >= m + 1) Init_U(x << 1 | 1, m + 1, t, Pos);
}
 
void Turn(int x, int s, int t, int Pos, int f)
{
    if (Pos >= t)
    {
        if (C[x] == Used_Index) return;
        C[x] = Used_Index;
        U[x] = Son[U[x]][f];
        return;
    }
    int m = (s + t) >> 1;
    Turn(x << 1, s, m, Pos, f);
    if (Pos >= m + 1) Turn(x << 1 | 1, m + 1, t, Pos, f);
}
 
int Get_Sum(int x, int s, int t, int Pos)
{
    if (Pos >= t) return T[Son[U[x]][0]];
    int ret = 0, m = (s + t) >> 1;
    ret += Get_Sum(x << 1, s, m, Pos);
    if (Pos >= m + 1) ret += Get_Sum(x << 1 | 1, m + 1, t, Pos);
    return ret;
}
 
int main() 
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &A[i]);
        Change(1, 0, n, i, A[i], 1);
    }
    char f;
    int L, R, Pos, Num, k;
    for (int i = 1; i <= m; ++i)
    {
        f = '-';
        while (f < 'A' || f > 'Z') f = getchar();
        if (f == 'C')
        {
            scanf("%d%d", &Pos, &Num);
            Change(1, 0, n, Pos, A[Pos], -1);
            A[Pos] = Num;
            Change(1, 0, n, Pos, Num, 1);
        }  
        else
        {
            scanf("%d%d%d", &L, &R, &k);
            int l, r, mid, Temp;
            Used_Index = 0;
            Init_U(1, 0, n, L - 1);
            Init_U(1, 0, n, R);
            l = 0; r = MN;
            while (l < r)
            {
                ++Used_Index;
                mid = (l + r) >> 1;
                Temp = Get_Sum(1, 0, n, R) - Get_Sum(1, 0, n, L - 1);
                if (Temp >= k)
                {
                    r = mid;
                    Turn(1, 0, n, R, 0);
                    Turn(1, 0, n, L - 1, 0);
                }
                else
                {
                    l = mid + 1;
                    Turn(1, 0, n, R, 1);
                    Turn(1, 0, n, L - 1, 1);
                    k -= Temp;
                }
            }
            printf("%d\n", l);
        }
    }
    return 0;
}

　　

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