[Codeforces967C]Stairs and Elevators(二分查找)

[不稳定的传送门]

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);
	}
}