Problem G GentleBots (the use of structures is worth learning)

Topic link: https://vjudge.net/problem/Gym-101606G

题目：
Rainforest Inc. is opening a large new automated warehouse in the far Northern reaches of the
UK—some place they call “Walthamstow”.
The robotic worker drones inside will operate in not just one or two, but three dimensions, and
can move in any one of the 6 cardinal dimensions in steps of 1 metre at a time. For example,
a robot looking to move from position (X1, Y1, Z1) to position (X2, Y2, Z2) will, assuming no
obstacles, need to take (|X2 − X1| + |Y2 − Y1| + |Z2 − Z1|) steps in total.
Since this warehouse is in Britain, and because every stereotype is true, the robotic denizens are
all impeccably polite. When two robots travelling in opposite directions meet, they wordlessly
negotiate for one of the robots to step aside somehow so the other can pass.
Multiple robots cannot occupy the same integer co-ordinates, and no two robots can swap
positions in one move. All moves are instantaneous.
We have prepared a test run of the warehouse with just two machines installed. Write a program
to pass this test.
Input
Two lines, one for each robot, each containing six space-separated integers (X0Y0Z0) and
(X∞Y∞Z∞), the intended start and end locations of a robot respectively (−1000 ≤ X, Y, Z ≤
1000).
The robots will start in different places from each other, and will also end in different places
from each other.
Output
Output up to 7000 lines, giving one possible list of locations of the robots over time. The position
of both robots at time T must be given as bracketed space-separated (X, Y, Z) co-ordinate tuples
on line T.
Co-ordinates must not exceed an absolute value of 106
.
Sample Input 1 Sample Output 1
0 0 0 2 2 2
1 1 2 0 0 0
(0 0 0) (1 1 2)
(1 0 0) (1 1 1)
(1 1 0) (0 1 1)
(1 1 1) (0 1 0)
(1 1 2) (0 0 0)
(1 2 2) (0 0 0)
(2 2 2) (0 0 0)
Sample Input 2 Sample Output 2
-2 0 0 1 0 0
3 0 0 -1 0 0
(-2 0 0) (3 0 0)
(-1 0 0) (2 0 0)
(0 0 0) (1 0 0)
(1 0 0) (1 0 -1)
(1 0 0) (0 0 -1)
(1 0 0) (-1 0 -1)
(1 0 0) (-1 0 0)
Sample Input 3 Sample Output 3
0 0 0 1 0 0
1 0 0 0 0 0
(0 0 0) (1 0 0)
(0 1 0) (0 0 0)
(1 1 0) (0 0 0)
(1 0 0) (0 0 0

The meaning of the question: In a three-dimensional coordinate system, you are given the start and end coordinates of two robots. Each robot can only move one unit in the x or y or z direction within 1s or stay still, allowing you to output every second. For the coordinates of the two robots, there are many answers, just output any one. Both robots are moving at the same time, they cannot be on the same coordinate at the same time, and the two robots cannot move alternately (one robot leaves this coordinate, the other robot enters this coordinate).

Ideas: Let the evaluation function be the sum of the Manhattan distances from two points to the end point, and choose the direction that makes the evaluation function optimal each time. If it cannot go, it will randomly shake.

The code is transferred from a big guy : https://www.cnblogs.com/clrs97/p/7768748.html

#include<cstdio>
#include<ctime>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int dx[7]={1,-1,0,0,0,0,0},
          dy[7]={0,0,-1,1,0,0,0},
          dz[7]={0,0,0,0,1,-1,0};
struct P{
    int x,y,z;
    void read(){
        scanf("%d%d%d",&x,&y,&z);
    }
    int dis(P b){
        return abs(x-b.x)+abs(y-b.y)+abs(z-b.z);
    }
    void write(){
        printf("(%d %d %d)",x,y,z);
    }
    P(){}
    P(int _x,int _y,int _z){x=_x,y=_y,z=_z;}
    P apply(int d){
        return P(x+dx[d],y+dy[d],z+dz[d]);
    }
}A,B,C,D;//A->B C->D
int main(){
    A.read();
    B.read();
    C.read();
    D.read();
    while(1){
        A.write();
        putchar(' ');
        C.write();
        puts("");
        int pre=A.dis(B)+C.dis(D);
        if(!pre)return 0;
        int best=~0U>>1;
        int I=0,J=0;
        for(int i=0;i<7;i++)for(int j=0;j<7;j++){
            P NA=A.apply(i),NC=C.apply(j);
            if(!NA.dis(C))continue;
            if(!NA.dis(NC))continue;
            if(!NC.dis(A))continue;
           // if(!NC.dis(NA))continue;
            int now=NA.dis(B)+NC.dis(D);
            if(now<best)best=now,I=i,J=j;
        }
        if(best>=pre){
            while(1){
                int i=rand()%7,j=rand()%7;
                P NA=A.apply(i),NC=C.apply(j);
                if(!NA.dis(C))continue;
                if(!NA.dis(NC))continue;
                if(!NC.dis(A))continue;
               // if(!NC.dis(NA))continue;
                I=i,J=j;
                break;
            }
        }
        A=A.apply(I);
        C=C.apply(J);
    }
}