【bzoj3524/2223】[Poi2014]Couriers/[Coci 2009]PATULJCI 主席树

Description

给一个长度为n的序列a。1≤a[i]≤n。
m组询问，每次询问一个区间[l,r]，是否存在一个数在[l,r]中出现的次数大于(r-l+1)/2。如果存在，输出这个数，否则输出0。

Input

第一行两个数n，m。
第二行n个数，a[i]。
接下来m行，每行两个数l,r，表示询问[l,r]这个区间。

Output

m行，每行对应一个答案。

Sample Input

7 5

1 1 3 2 3 4 3

1 3

1 4

3 7

1 7

6 6

Sample Output

1

0

3

0

4

HINT

【数据范围】

n,m≤500000

Source

By Dzy

建出主席树之后区间内左右看看那边大于len/2就好了…

本傻X再也不写指针版的了…卡空间卡的死死的…1kw的数组才够用啊…你萌猜猜我RE几次…

顺便吐槽一下bzoj2223的读入好鬼畜

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

typedef long long LL;
const int SZ = 500010;
const int INF = 1000000010;
const double eps = 1e-6;

int n,m;

struct node{
    int l,r;
    int cnt;
}tree[10000010];

int rt[SZ];

int Tcnt = 0;

void insert(int l,int r,int last,int &now,int x)
{
    now = ++ Tcnt;
    tree[now] = tree[last];
    tree[now].cnt = tree[last].cnt + 1;
    if(l == r) return;
    int mid = l + r >> 1;
    if(x <= mid) insert(l,mid,tree[last].l,tree[now].l,x);
    else insert(mid + 1,r,tree[last].r,tree[now].r,x);
}

int ask(int l,int r)
{
    int len = r - l + 1 >> 1;
    int p = rt[l - 1];
    int q = rt[r];
    int s = 1,t = n;
    if(tree[q].cnt - tree[p].cnt <= len) return 0;
    while(s != t)
    {
        int mid = s + t >> 1;
        if(tree[tree[q].l].cnt - tree[tree[p].l].cnt > len)
            q = tree[q].l,p = tree[p].l,t = mid;
        else if(tree[tree[q].r].cnt - tree[tree[p].r].cnt > len)
            q = tree[q].r,p = tree[p].r,s = mid + 1;
        else
            return 0;
    }
    return s;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n;i ++)
    {
        int x;
        scanf("%d",&x);
        insert(1,n,rt[i - 1],rt[i],x);
    }
    for(int i = 1;i <= m;i ++)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        printf("%d\n",ask(l,r));
    }
    return 0;
}

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

typedef long long LL;
const int SZ = 500010;
const int INF = 1000000010;
const double eps = 1e-6;

int n,m,L;

struct node{
    int l,r;
    int cnt;
}tree[10000010];

int rt[SZ];

int Tcnt = 0;

void insert(int l,int r,int last,int &now,int x)
{
    now = ++ Tcnt;
    tree[now] = tree[last];
    tree[now].cnt = tree[last].cnt + 1;
    if(l == r) return;
    int mid = l + r >> 1;
    if(x <= mid) insert(l,mid,tree[last].l,tree[now].l,x);
    else insert(mid + 1,r,tree[last].r,tree[now].r,x);
}

int ask(int l,int r)
{
    int len = r - l + 1 >> 1;
    int p = rt[l - 1];
    int q = rt[r];
    int s = 1,t = L;
    if(tree[q].cnt - tree[p].cnt <= len) return 0;
    while(s != t)
    {
        int mid = s + t >> 1;
        if(tree[tree[q].l].cnt - tree[tree[p].l].cnt > len)
            q = tree[q].l,p = tree[p].l,t = mid;
        else if(tree[tree[q].r].cnt - tree[tree[p].r].cnt > len)
            q = tree[q].r,p = tree[p].r,s = mid + 1;
        else
            return 0;
    }
    return s;
}

int main()
{
    scanf("%d%d",&n,&L);
    for(int i = 1;i <= n;i ++)
    {
        int x;
        scanf("%d",&x);
        insert(1,L,rt[i - 1],rt[i],x);
    }
    int q;
    scanf("%d",&q);
    for(int i = 1;i <= q;i ++)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        int ans = ask(l,r);
        if(ans == 0) puts("no");
        else printf("yes %d\n",ans);
    }
    return 0;
}