2021 Niu Ke Winter Holiday Algorithm Basic Training Camp 3 E-Buy Gifts (the segment tree interval is the largest)

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Topic

• n cabinets, each cabinet’s gift number is , q times inquired.
• 1 x, which means removing the gift from the x cabinet. 2.     Indicates whether there are two gifts with the same number in the query interval [xi,yi].
• For each query output 0/1.

Ideas

• At first glance, it is a data structure problem, but it is also a thinking problem.
• For each cabinet i, pre[i] represents the subscript of the nearest cabinet with the same gift number as ai before i, and pos[i] represents the subscript of the nearest cabinet with the same gift number as ai after i.
• To construct a line segment tree for pre[i], you only need to query the maximum value of the interval when querying. If the maximum value of this interval >= , it means that there is a gift in the interval and the previous position is also in the interval. If the condition is met, output 1; otherwise, output 0.
• Then there is the interval modification . If the cabinet x withdraws the gift, then the value in the line segment tree should not work, so we can directly change it to 0 , and it will not affect others. At the same time, we need to update the cabinet j of pre[x] so that pos[j] is no longer x, but pos[x] , which is equivalent to a linked list deletion operation.

ac code

#include<bits/stdc++.h>
using namespace std;
#define io cin.tie(0);ios::sync_with_stdio(false);
#define debug(x) cout<<#x<<"="<<x<<endl
#define lowbit(x) x&(-x)
#define pii pair<int,int>
#define mk make_pair
#define ll long long
#define rs p<<1|1
#define ls p<<1
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
inline ll read(){
    ll p=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){p=(p<<1)+(p<<3)+(c^48),c=getchar();}
    return f*p;
}

struct node{
    int l, r;
    int lazy, sum, mx;
}tree[maxn << 2];
int a[maxn], pre[maxn], pos[maxn];
void pushup(int p){
    tree[p].mx = max(tree[ls].mx, tree[rs].mx);
}

void build(int p, int l, int r){
    tree[p].l = l; tree[p].r = r;
    if(l == r){
        tree[p].mx = pre[l];
        return;
    }
    int mid = l + r >> 1;
    build(ls, l, mid);
    build(rs, mid + 1, r);
    pushup(p);
}
void update(int p, int l, int r, int val) {//区间的每个值都改为val
    if(l <= tree[p].l && tree[p].r <= r){
        tree[p].mx = val;
        return;
    }
    int mid = tree[p].l + tree[p].r >> 1;
    if(mid >= l) update(ls, l, r, val);
    if(mid + 1 <= r) update(rs, l, r, val);
    pushup(p);
}
int query(int p, int l, int r){
    if(l <= tree[p].l && tree[p].r <= r)
        return tree[p].mx;
    int ans = -inf;
    int mid = tree[p].l + tree[p].r >> 1;
    if(mid >= l) ans = max(ans, query(ls, l, r));
    if(mid + 1 <= r) ans = max(ans, query(rs, l, r));
    return ans;
}
void solve(){
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i ++){
        a[i] = pre[i] = 0;
        pos[i]= n + 1;
    }
    for(int i = 1; i <= n; i ++){
        int x; scanf("%d", &x);
        pos[a[x]] = i;
        pre[i] = a[x];
        a[x] = i; //当前x出现的最近位置
    }
    build(1, 1, n);
    while(m --){
        int op, x;
        scanf("%d", &op);
        if(op == 1){
            scanf("%d", &x);
            update(1, x, x, 0);
            if(pos[x] <= n){
                update(1, pos[x], pos[x], pre[x]); //如果礼物ax后面还有的话更新
                pos[pre[x]] = pos[x];
            }
        }else{
            int l, r;
            scanf("%d%d", &l, &r);
            printf("%d\n", query(1, l, r) >= l);
        }
    }
}
int main(){
    solve();
    return 0;
}