分块与树状数组：两种解法实现数据操作-优快云博客

本文链接：https://blog.youkuaiyun.com/Protinx/article/details/126790984

本文介绍了如何使用分块思想和树状数组算法解决数据操作问题，包括快速查找中位数、元素的插入与删除。通过实例展示了这两种数据结构在处理查询和维护中位数问题上的高效性。

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分块思想解法
#include <cmath>
#include <cstdio>
#include <cstring>
#include <stack>
using namespace std;
const int maxn = 100010;
const int sqrN = 316;
stack<int> st;
int block[sqrN];
int table[maxn];
void peekMedian(int K) {
  int sum = 0;
  int idx = 0;
  while (sum + block[idx] < K) {
    sum += block[idx++];
  }
  int num = idx * sqrN;
  while (sum + table[num] < K) {
    sum += table[num++];
  }
  printf("%d\n", num);
}
void Push(int x) {
  st.push(x);
  block[x / sqrN]++;
  table[x]++;
}
void Pop() {
  int x = st.top();
  st.pop();
  block[x / sqrN]--;
  table[x]--;
  printf("%d\n", x);
}
int main() { 
    int x, query; 
    memset(block,0,sizeof(block));
    memset(table, 0, sizeof(table));
    char cmd[20];
    scanf("%d",&query);
    for(int i=0;i<query;i++){
        scanf("%s",cmd);
        if(strcmp(cmd,"Push")==0){
            scanf("%d",&x);
            Push(x);
        }else if(strcmp(cmd,"Pop")==0){
            if(st.empty()==true){
                printf("Invalid\n");
            }else{
                Pop();
            }
        }else{
            if(st.empty()==true){
                printf("Invaild\n");
            }else {
            int K=st.size();
            if(K%2==1)K=(K+1)/2;
            else K=K/2;
            peekMedian(K);
            }
    }
}
return 0;
}
树状数组解法
#include <cmath>
#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
#define lowbit(i) ((i)&-(i))
const int MAXN=100010;
stack<int>s;
int c[MAXN];
void update(int x,int v){
    for(int i=x;i<MAXN;i+=lowbit(i)){
        c[i]+=v;
    }
}
int getSum(int x){
    int sum=0;
    for(int i=x;i>0;i-=lowbit(i)){
        sum+=c[i];
    }
    return sum;
}
void PeekMedian(){
    int l=1,r=MAXN,mid,K=(s.size()+1)/2;
    while(l<r){
        mid=(1+r)/2;
        if(getSum(mid)>=K)r=mid;
        else l=mid+1;
    }
    printf("%d\n",l);
}
int main(){
    int n,x;
    char str[12];
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%s",str);
        if(strcmp(str,"Push")==0){
            scanf("%d",&x);
            s.push(x);
            update(x,1);
        }else if(strcmp(str,"Pop")==0){
            if(s.empty()) printf("Invaild\n");
            else
            {
                printf("%d\n",s.top());
                update(s.top(),-1);
                s.pop();
            }
        }else if(strcmp(str,"PeekMedian")==0){
            if(s.empty())printf("Invaild\n");
            else PeekMedian();
        }
    }
    return 0;
}