Sloution
每次试一下最近的2个楼梯或者电梯就行了
Code
#include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; int n,m,cl,q,ce,v,l[100010],e[100010],X1,X2,Y1,Y2,Ans,p; inline int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int main(){ n=read(),m=read(),cl=read(),ce=read(),v=read(); for(int i=1;i<=cl;++i) l[i]=read(); for(int i=1;i<=ce;++i) e[i]=read(); q=read(); while(q--){ X1=read(),Y1=read(),X2=read(),Y2=read();Ans=0x7fffffff; if(X1==X2) Ans=fabs(Y1-Y2); else{ p=lower_bound(l+1,l+cl+1,Y1)-l; if(p<=cl) Ans=min(Ans,(int)fabs(l[p]-Y1)+(int)fabs(X1-X2)+(int)fabs(l[p]-Y2)); if(--p) Ans=min(Ans,(int)fabs(l[p]-Y1)+(int)fabs(X1-X2)+(int)fabs(l[p]-Y2)); p=lower_bound(e+1,e+ce+1,Y1)-e; if(p<=ce) Ans=min(Ans,(int)fabs(e[p]-Y1)+((int)fabs(X1-X2)+v-1)/v+(int)fabs(e[p]-Y2)); if(--p) Ans=min(Ans,(int)fabs(e[p]-Y1)+((int)fabs(X1-X2)+v-1)/v+(int)fabs(e[p]-Y2)); } printf("%d\n",Ans); } }