傳送門

細節題……

首先離散化。

然後用線段樹維護最大值

因為左端點和右端點的存在性，要做一系列的討論。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 50005


struct Node {
    int year,val;
}G[MAXN];

int N,M;
bool cmp(Node a,Node b) {return a.year<b.year;}

struct disc {
    void work() {
        std::sort(G+1,G+N+1,cmp);
    }
    int num(int x,int opt) {
        int l = 1,r = N;
        while(l<r) {
            int mid = (l+r)>>1;
            if(G[mid].year>=x) r = mid;
            else l = mid + 1;
        }
        if(opt==2&&G[l].year!=x) return l-1;
        return l;
    }
}d;

#define lson (rt<<1)
#define rson (rt<<1|1)
#define mid ((l+r)>>1)

int seg[MAXN<<2];

inline void pushup(int rt) {
    seg[rt] = std::max(seg[lson],seg[rson]);
}

void Build(int rt,int l,int r) {
    if(l==r) {
        seg[rt] = G[l].val;
        return;
    }
    Build(lson,l,mid);
    Build(rson,mid+1,r);
    pushup(rt);
}

int query(int L,int R,int rt,int l,int r) {
    if(L>r||R<l||L>R) return 0;
    if(L<=l&&R>=r) return seg[rt];
    int ans = 0;
    if(L<=mid) ans = query(L,R,lson,l,mid);
    if(R>mid) ans = std::max(ans,query(L,R,rson,mid+1,r));
    return ans;
}

int main() {
    
    scanf("%d",&N);
    for(int i=1;i<=N;++i) scanf("%d%d",&G[i].year,&G[i].val);
    d.work(); Build(1,1,N);
    
    int x,y;
    scanf("%d",&M);
    for(int i=1;i<=M;++i) {
        
        scanf("%d%d",&x,&y);
        int l = d.num(x,1),r = d.num(y,2);
        
        int t1 = query(l,l,1,1,N);
        int t3 = query(r,r,1,1,N);

        bool f1 = false,f2 = false;

        if(G[l].year==x) ++l,f1 = true;
        if(G[r].year==y) --r,f2 = true;
        
        int t2 = query(l,r,1,1,N);

        if(f1) --l;
        if(f2) ++r;        

        if(x==y) puts("true");
        else if(G[l].year!=x&&G[r].year!=y) puts("maybe");
        else if(G[l].year!=x) {
            if(x==y-1) puts("maybe");
            else {
                if(t2>=t3) puts("false");
                else puts("maybe");
            }
        
        }
        else if(G[r].year!=y) {
            if(x==y-1) puts("maybe");
            else {
                if(t2>=t1) puts("false");
                else puts("maybe");
            }
        }
        else {
            if(t3>t1) puts("false");
            else if(l==r-1) {
                if(x==y-1) puts("true");
                else puts("maybe");
            }
            else {
                if(t2>=t3) puts("false");
                else if(r-l==y-x) puts("true");
                else puts("maybe");
            }
        }     
    }
    return 0;
}