Educational Codeforces Round 6 A. Professor GukiZ's Robot 水

In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its y coordinate and get the finish position.

In the second example robot should simultaneously increase x coordinate and decrease y coordinate by one three times.

题意：

　　给你一个起点和一个终点，robot可以向自己所在起点的周围8个方向走，问你最短路？

题解：

　　答案就是max(abs(x2-x1),abs(y1-y2)) 

#include <cstdio>
#include <cstring>
#include <ctime>
#include <algorithm>
using namespace std;
typedef long long ll;
int x1,x2,y2,y1;
int main() {
    scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
    printf("%d\n",max(abs(x2-x1),abs(y1-y2)) );
    return 0;
}