题意：n*m的楼房，cl个楼梯，ce个电梯，除了电梯的最大速度是v外，其他速度都是1。给出q次询问，回答(x1, y1)到(x2, y2)，最少需要多少时间？
顺序输入（有小到大）；

简短的。。。。。

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

const int maxn = 1e5+5;
int l[maxn];
int e[maxn];


int main(){
int n,m,ce,cl,q,x1,x2,yy1,yy2,v,ans;
    while(cin>>n>>m>>cl>>ce>>v){

        for(int i=0;i<cl;i++) cin>>l[i];
        for(int i=0;i<ce;i++) cin>>e[i];

        cin>>q;
        while(q--){
            ans=0x3f3f3f3f;
            cin>>yy1>>x1>>yy2>>x2;

            if(yy1==yy2){cout<<abs(x1-x2)<<endl;continue;}

            int lt=lower_bound(l,l+cl,x1)-l;
            int dt=lower_bound(e,e+ce,x1)-e;

            if(lt<cl) ans=min(ans,abs(l[lt]-x1)+abs(l[lt]-x2)+abs(yy2-yy1));
            if(lt>0) ans=min(ans,abs(l[lt-1]-x1)+abs(l[lt-1]-x2)+abs(yy2-yy1));

            int sx=(abs(yy1-yy2)+v-1)/v;

            if(dt<ce) ans=min(ans,(int)(abs(e[dt]-x1)+abs(e[dt]-x2)+sx));
            if(dt>0)  ans=min(ans,(int)(abs(e[dt-1]-x1)+abs(e[dt-1]-x2)+sx));

            printf("%d\n",ans);
        }
    }
    return 0;
}

我最开始写的 orz（脑子抽了写这干什么）。。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <list>
#include <bitset>
using namespace std;
typedef long long ll;

const int INF=0x3f3f3f3f;
const int maxn=1e5+5;
int dianti[maxn];
int louti[maxn];
int n,m,c1,c2,v,q;

int main(){
    while(cin>>n>>m>>c1>>c2>>v){
    // memset(dianti)
        for(int i=0;i<c1;i++) scanf("%d",&louti[i]);
        for(int i=0;i<c2;i++) scanf("%d",&dianti[i]);
        scanf("%d",&q);
        int x1,x2,y1,y2;
        for(int i=0;i<q;i++){
            cin>>y1>>x1>>y2>>x2;
            if(y1==y2){cout<<abs(x2-x1)<<endl;continue;}
            int ans1=INF,ans2=INF;
            if(c1!=0){
                int louti_r_pos=upper_bound(louti,louti+c1,x1)-louti;
                int louti_l_pos=-1;

                int louti_r;
                if(louti_r_pos==c1){
                    louti_r=louti[louti_r_pos-1];
                    louti_r_pos--;
                }
                else{
                    louti_r=louti[louti_r_pos];
                }
                if(louti_r_pos-1>=0) louti_l_pos=louti_r_pos-1;
                int louti_l=-1;
                if(louti_l_pos>=0) louti_l=louti[louti_l_pos];

                int louti_y_time=abs((y2-y1));

                int louti_rx_time=abs(x1-louti_r);
                int louti_lx_time=louti_l!=-1?abs(x1-louti_l):INF;

                int louti_r_tsum=louti_y_time+louti_rx_time+abs(x2-louti_r);
                int louti_l_tsum=louti_y_time+louti_lx_time+abs(x2-louti_l);

                ans1=min(louti_r_tsum,louti_l_tsum);
            // cout<<" "<<ans1<<endl;
            }
            if(c2!=0){
                int dianti_r_pos=upper_bound(dianti,dianti+c2,x1)-dianti;

                int dianti_l_pos=-1;

                int dianti_r;
                if(dianti_r_pos==c2){
                    dianti_r=dianti[dianti_r_pos-1];
                    dianti_r_pos--;
                }
                else{
                    dianti_r=dianti[dianti_r_pos];
                }

                if(dianti_r_pos-1>=0) dianti_l_pos=dianti_r_pos-1;
                int dianti_l=-1;    
                if(dianti_l_pos>=0) dianti_l=dianti[dianti_l_pos];  

                int dianti_rx_time=abs(x1-dianti_r);
                int dianti_lx_time=dianti_l_pos!=-1?abs(x1-dianti_l):INF;

                int dianti_y_time=ceil(abs(1.0*y1-1.0*y2)/v);

                int dianti_r_tsum=dianti_rx_time+dianti_y_time+abs(x2-dianti_r);
                int dianti_l_tsum=dianti_lx_time+dianti_y_time+abs(x2-dianti_l);

                ans2=min(dianti_r_tsum,dianti_l_tsum);
            // cout<<" "<<ans2<<endl;
            }           
            printf("%d\n",(min(ans1,ans2)));
        }
    }
    return 0;
}