bzoj 2653middle

Description

一个长度为n的序列a，设其排过序之后为b，其中位数定义为b[n/2]，其中a,b从0开始标号,除法取下整。给你一个

长度为n的序列s。回答Q个这样的询问：s的左端点在[a,b]之间,右端点在[c,d]之间的子序列中，最大的中位数。

其中a<b<c<d。位置也从0开始标号。我会使用一些方式强制你在线。

Input

第一行序列长度n。接下来n行按顺序给出a中的数。

接下来一行Q。然后Q行每行a,b,c,d，我们令上个询问的答案是

x(如果这是第一个询问则x=0)。

令数组q={(a+x)%n,(b+x)%n,(c+x)%n,(d+x)%n}。

将q从小到大排序之后，令真正的

要询问的a=q[0],b=q[1],c=q[2],d=q[3]。　　

输入保证满足条件。

第一行所谓“排过序”指的是从小到大排序！

n<=20000,Q<=25000

Sample Output

271451044
271451044
969056313

我线段树查询某端最大值时写错了，我是直接+nd->ls->sum, 其实应该查询（L, R, Ls）的sum，因为L不一定在mid的左边；这个沙雕问题调了好久啊

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
const int M = 200005;
struct Node{
    Node *ls, *rs;
    int sum, zd[2];
    void up(){
        sum = ls->sum + rs->sum;
        zd[0] = max(ls->sum + rs->zd[0], ls->zd[0]);
        zd[1] = max(rs->sum + ls->zd[1], rs->zd[1]);
        
    }
}pool[M * 32], *tail = pool, *zero, *root[M];
int a[M];
struct Point{int pos, val;}b[M];
bool cmp(Point a, Point b){
    return a.val < b.val;
}
int n;
#define Ls nd->ls, lf, mid
#define Rs nd->rs, mid + 1, rg
int query(int L, int R, int opt, Node * nd, int lf = 1, int rg = n){
    if(L > R)return 0;
    if(lf >= L && rg <= R){
        if(opt == 2)return nd->sum;
        return nd->zd[opt];
    }
    int mid = (lf + rg) >> 1;
    int sum = 0;
    if(opt == 2){
        if(L <= mid)sum += query(L, R, opt, Ls);
        if(R > mid)sum += query(L, R, opt, Rs);    
        return sum;
    }
    else if(opt == 0){
        if(R > mid)sum = query(L, R, opt, Rs);
        if(L <= mid)sum = max(query(L, R, 2, Ls) + sum, query(L, R, opt, Ls));
            
    }
    else {
        if(L <= mid)sum = query(L, R, opt, Ls);
        if(R > mid)sum = max(query(L, R, 2, Rs) + sum , query(L, R, opt, Rs));
            
    }
    return max(sum, 0);
}
int x;

Node * build(int lf = 1, int rg = n){
    Node *nd = ++tail; 
    if(lf == rg) nd->sum = nd->zd[0] = nd->zd[1] = 1;
    else {
        int mid = (lf + rg) >> 1;
        nd->ls = build(lf, mid);
        nd->rs = build(mid + 1, rg);
        nd->up();
    }
    return nd;
}
Node * insert(int pos, Node *nd, int lf = 1, int rg = n){
    Node *nnd = ++tail;
    if(lf == rg)nnd->sum = nnd->zd[0] = nnd->zd[1] = -1;
    else {
        int mid = (lf + rg) >> 1;
        if(pos <= mid){
            nnd->rs = nd->rs;
            nnd->ls = insert(pos, Ls);
        }
        else {
            nnd->ls = nd->ls;
            nnd->rs = insert(pos, Rs);
        }
        nnd->up();
    }
    return nnd;
}
int q[5];
int Query(){
    int ans = 0;
    for(int i = 0; i < 4; i++)q[i]++;//, printf("%d ", q[i]);puts("");
    int lf = 1, rg = n;
    while(lf <= rg){
        int mid = (lf + rg) >> 1;
        int t1 = query(q[1], q[2], 2, root[mid]);
        int t2 = query(q[0], q[1] - 1, 1, root[mid]);
        int t3 = query(q[2] + 1, q[3], 0, root[mid]);
        int tt = t1 + t2 + t3;
        //printf("%d %d %d %d %d\n", mid, t1, t2, t3, tt);
        if(tt >= 0)ans = mid, lf = mid + 1;
        else rg = mid - 1;
    }
    //printf("DONE\n");
    return b[ans].val;
}


int main(){
//    freopen("1.in","r",stdin);
//    freopen("my.out","w",stdout);
    int m;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
        b[i].pos = i, b[i].val = a[i];
    }
    sort(b + 1, b + 1 + n, cmp);
    root[1] = build();
    //cout<<root[1]->sum<<endl;
    for(int i = 2; i <= n; i++){
        //b[i].id = i; 
        root[i] = insert(b[i - 1].pos, root[i - 1]);
        //cout<<root[i]->sum<<endl;
    }
    scanf("%d", &m);
    while(m--){
        int a, b, c, d;
        scanf("%d%d%d%d", &a, &b, &c, &d);
        q[0] = (a+x)%n, q[1] = (b+x)%n, q[2] = (c+x)%n, q[3] = (d+x)%n;
        sort(q, q + 4);
        x = Query();
        printf("%d\n", x);
    }    
    
}

View Code

原文地址：https://www.cnblogs.com/EdSheeran/p/9502130.html

bzoj 2653middle

2653: middle

Description

Input

Output

Sample Input

Sample Output