Description

Solution

看了题解才知道啥是ISIJ，infleaking好强啊%%%

记a[i]为从i往右走不换座位能走到的最右端。我们离线然后合并同一座位相交的线段，用线段树维护一下这个a
可以发现i向a[i]连边组成了一棵树，于是问题变成求两个点的深度差，这个用倍增做就行了

Code

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#define fill(x,t) memset(x,t,sizeof(x))
#define rep(i,st,ed) for (register int i=st;i<=ed;++i)
#define drp(i,st,ed) for (register int i=st;i>=ed;--i)
#define fi first
#define se second

typedef std:: pair <int,int> pair;
const int N=200005;

std:: vector <pair> vec[N];

int max[N<<2],tag[N<<2];
int fa[19][N];

int read() {
    int x=0,v=1; char ch=getchar();
    for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):(v),ch=getchar());
    for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
    return x*v;
}

void cmax(int &x,int y) {
    (y>x)?(x=y):(0);
}

void push_down(int now) {
    if (!tag[now]) return ;
    int w=tag[now]; tag[now]=0;
    cmax(tag[now<<1],w); cmax(tag[now<<1|1],w);
    cmax(max[now<<1],w); cmax(max[now<<1|1],w);
}

void modify(int now,int tl,int tr,int l,int r,int v) {
    if (tl==l&&tr==r) {
        cmax(tag[now],v);
        cmax(max[now],v);
        return ;
    }
    push_down(now);
    int mid=(tl+tr)>>1;
    if (r<=mid) modify(now<<1,tl,mid,l,r,v);
    else if (l>mid) modify(now<<1|1,mid+1,tr,l,r,v);
    else {
        modify(now<<1,tl,mid,l,mid,v);
        modify(now<<1|1,mid+1,tr,mid+1,r,v);
    }
    max[now]=std:: max(max[now<<1],max[now<<1|1]);
}

int query(int now,int tl,int tr,int x) {
    if (tl==tr) return max[now];
    int mid=(tl+tr)>>1;
    push_down(now);
    if (x<=mid) return query(now<<1,tl,mid,x);
    return query(now<<1|1,mid+1,tr,x);
}

int main(void) {
    freopen("trip.in","r",stdin);
    freopen("trip.out","w",stdout);
    int n=read(),m=read(),k=read();
    rep(i,1,m) {
        int s=read(),t=read()-1,a=read();
        vec[a].push_back(pair(s,t));
    }
    rep(i,1,k) {
        std:: sort(vec[i].begin(),vec[i].end());
        int L=0,R=-1;
        for (int j=0;j<vec[i].size();++j) {
            if (vec[i][j].fi<=R+1) {
                R=std:: max(R,vec[i][j].se);
                continue;
            }
            modify(1,0,n,R+1,vec[i][j].fi-1,vec[i][j].fi);
            L=vec[i][j].fi; R=vec[i][j].se;
        }
        modify(1,0,n,R+1,n,n);
    }
    rep(i,1,n-1) {
        int tmp=query(1,0,n,i);
        if (!tmp) tmp=n+1;
        fa[0][i]=tmp;
    }
    fa[0][n]=n; fa[0][n+1]=n+1;
    rep(j,1,18) drp(i,n+1,1) {
        fa[j][i]=fa[j-1][fa[j-1][i]];
    }
    for (int T=read();T--;) {
        int x=read(),y=read(),ans=0;
        drp(i,18,0) (fa[i][x]<y)?(x=fa[i][x],ans+=(1<<i)):0;
        x=fa[0][x]; ans++;
        if (x==n+1||x<y) puts("-1");
        else printf("%d\n", ans);
    }
    return 0;
}