【链接】 我是链接,点我呀:)
【题意】
【题解】
假设我们从(r,c)出发想要到固定某个点(i,j)的最短距离 我们设x0为向左走动的次数,y0为向右走动的次数 显然(j-c)=y0-x0 即y0 = (j-c) + x0 这里j-c是一个常数 也就是说当我们让x0最小的时候到达每一个点的对应的y0也一定是最优的,因为和x0成正相关 因此我们只要求出来(r,c)到所有点所用的x0的最小值就好了 会发现没个点到达相邻的点的花费要么就是1(往左走),要么就是0(往其他方向走) 符合01bfs的特点 (01bfs:如果向左走的话,那么就把新得到的状态加入到末尾,否则加入到队头,因为花费为0的话,和之前的dis值相同,依然是当前所有的情况里最低花费 (所以可以用来继续更新最小值,因此是放在队头,这样的话能始终维持队列从头到尾的dis值依次递增的顺序 (因此能保证每次尝试用来更新的状态都是当前状况下最优的状态,不会造成错解或者漏解【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static class Pair{
int x,y,d;
public Pair(int x,int y,int d){
this.x = x;
this.y = y;
this.d = d;
}
}
static int N = 2000;
static class Task{
int n,m,r,c,x,y;
String s[];
int dis[][];
int dx[]= {0,0,1,-1},dy[]= {1,-1,0,0};
Deque<Pair> queue;
public void solve(InputReader in,PrintWriter out) {
queue = new LinkedList<Pair>();
dis = new int[N+10][N+10];
s = new String[N+10];
n = in.nextInt();m = in.nextInt();
r = in.nextInt();c = in.nextInt();
x = in.nextInt();y = in.nextInt();
for (int i = 1;i <= n;i++) {
s[i] = in.next();
StringBuilder sb = new StringBuilder(s[i]);
sb.insert(0, ' ');
s[i] = sb.toString();
}
for (int i = 1;i <= N;i++)
for (int j = 1;j <= N;j++)
dis[i][j] = -1;
dis[r][c] = 0;
queue.offerFirst(new Pair(r,c,0));
while (!queue.isEmpty()) {
Pair temp = queue.pollFirst();
int x0 = temp.x,y0 = temp.y,d = temp.d;
for (int i = 0;i < 4;i++) {
int x1 = x0 + dx[i];int y1 = y0 + dy[i];
int td = d;
if (i==1) td++;
if (x1>=1 && x1<=n && y1>=1 && y1<=m && s[x1].charAt(y1)!='*') {
if (dis[x1][y1]==-1 ||dis[x1][y1]>td) {
dis[x1][y1] = td;
if (i==1) {
queue.offerLast(new Pair(x1,y1,td));
}else {
queue.offerFirst(new Pair(x1,y1,td));
}
}
}
}
}
int ans = 0;
for (int i = 1;i <= n;i++)
for (int j = 1;j <= m;j++) {
if (dis[i][j]==-1) continue;
int x0 = dis[i][j];
int y0 = j-c + x0;
if (x0<=x && y0<=y) {
ans++;
}
}
out.println(ans);
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}