C - Tile Distance 2

C - Tile Distance 2

https://atcoder.jp/contests/abc359/tasks/abc359_c

 

思路

在x方向上，让s<t

然后 如果s在tile的左边，移动到右边，  如果t在tile的右边，移动到左边，

计算x 和 y方便的必走的steps，

y方向上容易计算（跨的格子就是）， x方向有些复杂， s在x方向上，不用花费（配合y方向上走步），可延伸到最大位置为 sx+ysteps,

如果tx在此最大位置之内，则x方向不用花费， 否则必须花费。

 

t相对s三种位置。

 

 

Code

https://atcoder.jp/contests/abc359/submissions/54883741

long long sx, sy, tx, ty;int main()
{cin >> sx >> sy >> tx >> ty;/*keep source before target in x direction*/if (sx > tx){swap(sx, tx);swap(sy, ty);}/*if (sx,sy) in the left side of one bricki.e. sx+sy is even-------|s |  |-------then move s to its right tile, like below-------|  |s |-------*/if ((sx+sy)&1 == 0){sx++;}/*if (tx,ty) in the right side of one bricki.e. tx+ty is odd-------|  |s |-------then move s to its left tile, like below-------|s |  |-------*/if ((tx+ty)&1 == 1){tx--;}// xdiff must be not less than zero.// ydiff is of any valuelong long xdiff = tx - sx;long long ydiff = ty - sy;/*caculate the step of x-neccessary stepscaculate the step of y-neccessary steps*/long long xsteps = 0;long long ysteps = abs(ydiff);/*if x position of target is greater than sx extent scope,than x-neccessary steps are needed.*/long long sx_extent = sx + ysteps;if (tx > sx_extent){xsteps = (tx - sx_extent + 1)/2;}long long total = xsteps + ysteps;cout << total << endl;return 0;
}