Codeforces 877D Olya and Energy Drinks【思维优化Bfs】

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最新推荐文章于 2025-01-21 17:46:30 发布

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本文探讨了一个关于寻找从起点到终点最短时间的问题。主人公可以在一秒内向四个方向之一移动1到k米，需要避开障碍物。通过优化的广度优先搜索算法，解决了这一问题并提供了AC代码。

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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

input
3 4 4
....
###.
....
1 1 3 1

output
3

input
3 4 1
....
###.
....
1 1 3 1

output
8

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.

题目大意：

每秒主人公可以朝向一个方向走（1~k）步，问从起点走到终点最少需要多少秒。

思路：

直接Bfs的话肯定是水题。但是直接Bfs会TLE掉。我们考虑优化，对应我们如果走一条线的时候，如果出现了前方有一个位子（x，y），使得从起点走到这个点的时间小于当前点走到这个点的时间的话，无需继续向那个方向继续走了。因为那个方向的取优方案可以通过点（x，y）来更新，所以多余步骤可以删除。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
struct node
{
    int x,y,step;
}now,nex;
char a[1050][1050];
int dist[1050][1050];
int vis[1050][1050];
int n,m,k;
int sx,sy,ex,ey;
int fx[4]={0,0,1,-1};
int fy[4]={1,-1,0,0};
void Slove()
{
    sx--;sy--;ex--;ey--;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            dist[i][j]=0x3f3f3f3f;
            vis[i][j]=0;
        }
    }
    queue<node>s;
    now.x=sx,now.y=sy,now.step=0;
    dist[sx][sy]=0;
    s.push(now);
    while(!s.empty())
    {
        now=s.front();s.pop();
        vis[now.x][now.y]=0;
        for(int i=0;i<4;i++)
        {
            for(int j=1;j<=k;j++)
            {
                nex.x=now.x+fx[i]*j;
                nex.y=now.y+fy[i]*j;
                nex.step=now.step+1;
                if(nex.x<0||nex.x>=n||nex.y<0||nex.y>=m||a[nex.x][nex.y]=='#'||(dist[nex.x][nex.y]!=0x3f3f3f3f&&dist[nex.x][nex.y]<now.step+1))break;
                else
                {
                    if(nex.step<dist[nex.x][nex.y]&&vis[nex.x][nex.y]==0)
                    {
                        vis[nex.x][nex.y]=1;
                        dist[nex.x][nex.y]=nex.step;
                        s.push(nex);
                    }
                }
            }
        }
    }
    if(dist[ex][ey]==0x3f3f3f3f)dist[ex][ey]=-1;
    printf("%d\n",dist[ex][ey]);
}
int main()
{
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        for(int i=0;i<n;i++)scanf("%s",a[i]);
        scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
        Slove();
    }
}