[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; }