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题目链接:BZOJ - 1901
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 n)。
树状数组套线段树
#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;
}
[BZOJ 1901] Dynamic Rankings 【树状数组套线段树 || 线段树套线段树】
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原文地址:http://www.cnblogs.com/JoeFan/p/4317275.html