Codeforces 877 D. Olya and Energy Drinks (bfs)__搜尋

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Description

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

  Examples output

3

  題意

從 (x1,y1) (x_1,y_1) 走到 (x2,y2) (x_2,y_2) ,每次最多往一個方向走 k k 步,且不可穿牆,問最少需要走幾次才能達成目標。

  思路

每次把可以走到的點排入佇列 bfs 即可,注意最佳化剪枝,如同一個點只計算一次。

  AC 代碼

#include<bits/stdc++.h>#define IO ios::sync_with_stdio(false);\    cin.tie(0);\    cout.tie(0);#define inf 0x7f7f7f7fusing namespace std;typedef __int64 LL;const int maxm = 1e3+10;const int maxn = 1e7+10;const int mv[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};struct node{    int x;    int y;    int t;    node() {}    node(int xx,int yy,int tt)    {        x = xx;        y = yy;        t = tt;    }} g[maxn];char c[maxm][maxm];int f[maxm][maxm];int n,m,k;int x1,x2,y1,y2;queue<node>que;void init(){    for(int j=0; j<4; j++)        for(int i=1; i<=k; i++)        {            int xx = x1+mv[j][0]*i;            int yy = y1+mv[j][1]*i;            if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&c[xx][yy]=='.')            {                f[xx][yy]=1;                que.push(node(xx,yy,1));            }            else break;        }}void bfs(){    if(x1==x2&&y1==y2)    {        puts("0");        return;    }    init();    while(!que.empty())    {        node s=que.front();        que.pop();        if(f[x2][y2])        {            printf("%d\n",f[x2][y2]);            return;        }        for(int j=0; j<4; j++)            for(int i=1; i<=k; i++)            {                int xx = s.x+mv[j][0]*i;                int yy = s.y+mv[j][1]*i;                if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&c[xx][yy]=='.')                {                    if(f[xx][yy]==0)                    {                        f[xx][yy]=s.t+1;                        que.push(node(xx,yy,s.t+1));                    }                }                else break;            }    }    puts("-1");}int main(){    scanf("%d%d%d%*c",&n,&m,&k);    for(int i=1; i<=n; i++)        gets(c[i]+1);    scanf("%d%d%d%d",&x1,&y1,&x2,&y2);    bfs();    return 0;}
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