Description "Problem Background"
LHX has been plagued by many people who have come to worship him recently, so he added a barrier to his garden.
"Problem description"
The area near the leader's garden can be pumped into a network of square grids, each corresponding to a coordinate (all integers, possibly negative), if two meshes (x1, y1), (x2, y2) have |x1–x2| + |y1–y2| = 1, the two meshes are adjacent, otherwise they are not adjacent.
The leader sets a barrier on the grid at y = 0 for the entire line, which is a grid of all coordinates (x, 0). Of course, he also has to solve his own and internal personnel access problems, so the Bishop set n entrance A1, A2, ..., an accessible, that is, for all meshes on y = 0, only (A1, 0), (A2, 0), ..., (an, 0) can be passed, except that all grids with an ordinate of 0 cannot pass, And for a grid (x, y) with y not 0, it can be considered as random.
Now the leader wants to know that given m-point pairs (x1, y1), (x2, y2), and these points are not on the barrier, ask the shortest distance from one point to another point, each time only from one lattice to the adjacent lattice. the 1th behavior of input inputs is a positive integer n, which is the number of ingress on the barrier.
The 2nd line has n integers, a1, A2, ..., an, separated by a space for the horizontal axis of the N inlet.
The 3rd act is a positive integer m, which represents M queries.
Next m line, 4 integers per line x1, y1, x2, y2, y1 and y2 are not equal to 0, indicating a query from (x1, y1) to (x2, y2) the shortest path. The output outputs contain a total of M lines, and the first line of the output is the shortest distance from (x1, y1) to (x2, y2) of the I-query outputs. Sample Input
22-120 1 0-11 1 2 2
Sample Output
42
Hint "Data Size"
For 20% of the data, there is n,m≤10,ai,xi,yi absolute value of not more than 100;
For 40% of the data, there is n,m≤100,ai,xi,yi absolute value of not more than 1000;
For 60% of the data, there is n,m≤1000,ai,xi,yi absolute value of not more than 100000;
For 100% of the data, there is a n,m≤100000,ai,xi,yi absolute value of not more than 100000000. Exercises
- The main idea is that in a planar Cartesian coordinate system, the x-axis is a barrier, but there are n openings, which require two points of distance
- Obviously, this question can be discussed by category.
- ① the y-coordinate of the two points, directly seeking the Manhattan distance output
- If not, then two points in the middle of the opening
- ② if the opening is in the middle of two points, two points are added to the open Manhattan distance and output
- ③ Otherwise, determine which endpoint it is close to, and then go around it.
- Although the POS value at this point is not necessarily accurate, but very close, the real value must appear in the pos-1,pos,pos+1, to find the minimum output is good
Code
1#include <cstdio>2#include <iostream>3#include <algorithm>4 using namespacestd;5 intn,m,a[100010];6 intAbsintA) {returna<0?-a:a;}7 intFindintx)8 {9 intL=1, r=n,k=N;Ten while(l<=R) One { A intMid= (L+R)/2; - if(a[mid]<x) l=mid+1;Elser=mid-1, k=mid; - } the returnK; - } - intMain () - { + intX1,x2,y1,y2; -scanf"%d",&n); + for(intI=1; i<=n;i++) scanf ("%d",&a[i]); ASort (A +1, a+n+1); atscanf"%d",&m); - for(intI=1; i<=m;i++) - { -scanf"%d%d%d%d",&x1,&y1,&x2,&y2); - if((y1<0&&y2<0)|| (y1>0&&y2>0)) - { inprintf"%d\n", ABS (Y1-Y2) +abs (x1-x2)); - Continue; to } + if(x1>x2) Swap (X1,X2); - intPos=find ((X1+X2)/2); the if(a[pos]>=x1&&a[pos]<=x2) printf ("%d\n", ABS (Y1-Y2) +abs (x1-x2)); * Else $ {Panax Notoginseng intT1=abs (a[pos]-x1) +abs (a[pos]-x2) +abs (y1-y2), t2=0x7fffffff, t3=0x7fffffff; - if(pos>1) T2=abs (a[pos-1]-X1) +abs (a[pos-1]-X2) +abs (y1-y2); the if(pos<n) T3=abs (a[pos+1]-X1) +abs (a[pos+1]-X2) +abs (y1-y2); + if(T1>T2) t1=T2; A if(T1>T3) t1=T3; theprintf"%d\n", T1); + } - } $}
[dichotomy] [sort] Jzoj P1792 's garden
