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
");
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
", ans1, ans2, ans3);
}
}
return 0;
}