此题看似很简单，但实际上有不少细节，WA点不少。分情况处理即可。

#include<cmath>

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

bool inMap(int x, int y){

    return x > 0 && x < 9 && y > 0 && y < 9;

}

int main(){

    int r1, c1, r2, c2, ans1, ans2, ans3;

    //freopen("in.cpp", "r", stdin);

    while(~scanf("%d%d%d%d", &r1, &c1, &r2, &c2)){

        if(r1 == r2 && c1 == c2) printf("0 0 0\n");

        else{

            if(r1 == r2 || c1 == c2) ans1 = 1;

            else ans1 = 2;

            if(abs(r1 - r2) == abs(c1 - c2)) ans2 = 1;

            else if(r1 == r2 || c1 == c2){

                int tmp1 = abs(r1 - r2);

                int tmp2 = abs(c1 - c2);

                if((tmp1 % 2 == 0 && tmp1) || (tmp2 % 2 == 0 && tmp2)) ans2 = 2;

                else ans2 = 0;

            }

            else{

                int flag = 0, x[4], y[4];

                for(int i = 1;i < 8;i ++){

                    x[0] = r1 + i, y[0] = c1 + i;

                    x[1] = r1 - i, y[1] = c1 + i;

                    x[2] = r1 + i, y[2] = c1 - i;

                    x[3] = r1 - i, y[3] = c1 - i;

                    for(int j = 0;j < 4;j ++){

                        if(inMap(x[j], y[j]) && abs(x[j] - r2) == abs(y[j] - c2)){

                            flag = 1;

                            break;

                        }

                    }

                    if(flag) break;

                }

                if(flag) ans2 = 2;

                else ans2 = 0;

            }

            if(abs(r1 - r2) == abs(c1 - c2)) ans3 = abs(r1 - r2);

            else{

                int tmp1 = abs(r1 - r2) + abs(abs(c1 - c2) - abs(r1 - r2));

                int tmp2 = abs(c1 - c2) + abs(abs(r1 - r2) - abs(c1 - c2));

                ans3 = min(tmp1, tmp2);

            }

            printf("%d %d %d\n", ans1, ans2, ans3);

        }

    }

    return 0;

}