Codeforces Round #442 (Div. 2) D. Olya and Energy Drinks
D. Olya and Energy Drinks
Problem Statement
Olya loves energy drinks. She loves them so much that her room is full of empty cans from energy drinks.
Formally, her room can be represented as a field of n × m cells, each cell of which is empty or littered with cans.
Olya drank a lot of energy drink, so now she can run k meters per second. Each second she chooses one of the four directions (up, down, left or right) and runs from 1 to k meters in this direction. Of course, she can only run through empty cells.
Now Olya needs to get from cell ( x1 , y1 ) to cell ( x2 , y2 ). How many seconds will it take her if she moves optimally?
It’s guaranteed that cells ( x1 , y1 ) and ( x2 , y2 ) are empty. These cells can coincide.
Input
The first line contains three integers n, m and k (1 ≤ n, m, k ≤ 1000) — the sizes of the room and Olya’s speed.
Then n lines follow containing m characters each, the i-th of them contains on j-th position “#”, if the cell (i, j) is littered with cans, and “.” otherwise.
The last line contains four integers x1 , y1 , x2 , y2 (1 ≤ x1 , x2 ≤ n, 1 ≤ y1 , y2 ≤ m) — the coordinates of the first and the last cells.
Output
Print a single integer — the minimum time it will take Olya to get from ( x1 , y1 ) to ( x2 , y2 ).
If it’s impossible to get from ( x1 , y1 ) to ( x2 , y2 ), print -1.
Examples
Example 1
Input
3 4 4
. . . .
###.
. . . .
1 1 3 1
Output
3
Example 2
Input
3 4 1
. . . .
###.
. . . .
1 1 3 1
Output
8
Example 3
Input
2 2 1
. #
#.
1 1 2 2
Output
-1
Note
In the first sample Olya should run 3 meters to the right in the first second, 2 meters down in the second second and 3 meters to the left in the third second.
In second sample Olya should run to the right for 3 seconds, then down for 2 seconds and then to the left for 3 seconds.
Olya does not recommend drinking energy drinks and generally believes that this is bad.
题意
给一个n*m的迷宫，’#’代表不可走，’.’代表可走，给定一个k，每次可以向四个方向的其中一个方向直着走1~k步，然后给定起点和终点，问你最少走多少步就可以从起点走到终点。
思路
直接BFS就行了，虽然原来是n^3的BFS，肯定会T，不过只要多加几个剪枝，就可以跑得飞快了。。。具体剪枝看代码..
Code
#pragma GCC optimize(3)
#includeint ,int> >q;
int x,y,xx,yy,n,m,k,dist[1005][1005];
bool vis[1005][1005];
char mp[1005][1005];
int step[4][2]={{0,1},{1,0},{-1,0},{0,-1}};
int main() {
read(n);
read(m);
read(k);
for(int i=1;i<=n;i++) {
scanf("%s",mp[i]+1);
for(int j=1;j<=m;j++)
if(mp[i][j]=='#')
vis[i][j]=1;
}
read(x);
read(y);
read(xx);
read(yy);
memset(dist,0x3f3f3f3f,sizeof dist);
q.push(make_pair(x,y));
dist[x][y]=0;
while(!q.empty()) {
int nx=q.front().first,ny=q.front().second;
q.pop();
if(vis[nx][ny])
continue;
vis[nx][ny]=1;
for(int i=0;i<4;i++) {
int px=nx,py=ny;
for(int j=1;j<=k;j++) {
px+=step[i][0];
py+=step[i][1];
if(px<1||px>n||py<1||py>m)
break;
if(vis[px][py])
break;
if(dist[px][py]<=dist[nx][ny])
break;
if(dist[nx][ny]+11;
q.push(make_pair(px,py));
}
}
}
}
printf("%d\n",dist[xx][yy]==0x3f3f3f3f?-1:dist[xx][yy]);
return 0;
}