The pasture contains a small, contiguous grove of trees that has no ‘holes‘ in the middle of the it. Bessie wonders: how far is it to walk around that grove and get back to my starting position? She‘s just sure there is a way to do it by going from her start location to successive locations by walking horizontally, vertically, or diagonally and counting each move as a single step. Just looking at it, she doesn‘t think you could pass ‘through‘ the grove on a tricky diagonal. Your job is to calculate the minimum number of steps she must take. Happily, Bessie lives on a simple world where the pasture is represented by a grid with R rows and C columns (1 <= R <= 50, 1 <= C <= 50). Here‘s a typical example where ‘.‘ is pasture (which Bessie may traverse), ‘X‘ is the grove of trees, ‘*‘ represents Bessie‘s start and end position, and ‘+‘ marks one shortest path she can walk to circumnavigate the grove (i.e., the answer): ...+... ..+X+.. .+XXX+. ..+XXX+ ..+X..+ ...+++* The path shown is not the only possible shortest path; Bessie might have taken a diagonal step from her start position and achieved a similar length solution. Bessie is happy that she‘s starting ‘outside‘ the grove instead of in a sort of ‘harbor‘ that could complicate finding the best path.
牧场里有一片树林，林子里没有坑．
贝茜很想知道，最少需要多少步能围绕树林走一圈，最后回到起点．她能上下左右走，也能走对角线格子．牧场被分成R行C列(1≤R≤50，1≤C≤50)．下面是一张样例的地图，其中“．”表示贝茜可以走的空地， “X”表示树林， “*”表示起点．而贝茜走的最近的路已经特别地用“+”表示出来．
题目保证，最短的路径一定可以找到．
* Line 1: Two space-separated integers: R and C
* Lines 2..R+1: Line i+1 describes row i with C characters (with no spaces between them).
第1行输入R和C，接下来R行C列表示一张地图．地图中的符号如题干所述．
* Line 1: The single line contains a single integer which is the smallest number of steps required to circumnavigate the grove.
输出最少的步数．
#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
struct na{
int x,y;
};
int n,m,x=-1,y;
const int fx[8]={0,0,1,-1,1,1,-1,-1},fy[8]={1,-1,0,0,1,-1,1,-1};
int map[51][51];
char c;
queue <na> q;
int main(){
scanf("%d%d",&n,&m);
for (int i=0;i<n;i++)
for (int j=0;j<m;j++){
c=getchar();
while(c==‘\n‘) c=getchar();
if (c==‘X‘) {
map[i][j]=-1;
if (x==-1) x=i,y=j;
}else
if (c==‘*‘){
na tmp;tmp.x=i;tmp.y=j;q.push(tmp);
}else map[i][j]=2500;
}
while(!q.empty()){
na k=q.front();q.pop();
for (int i=0;i<8;i++){
na now=k;
now.x+=fx[i];now.y+=fy[i];
if (now.x<0||now.y<0||now.x>=n||now.y>=m) continue;
if (k.y<=y&&k.x==x&&now.x==x-1) continue;
if (k.y<=y&&k.x==x-1&&now.x==x) continue;
if (map[now.x][now.y]>map[k.x][k.y]+1){
map[now.x][now.y]=map[k.x][k.y]+1;
q.push(now);
}
}
}
int ans=2500;
for (int i=y-1;i>=0;i--){
if (map[x][i]+map[x-1][i]<ans) ans=map[x][i]+map[x-1][i];
if (map[x][i]+map[x-1][i+1]<ans) ans=map[x][i]+map[x-1][i+1];
if (i) if (map[x][i]+map[x-1][i-1]<ans) ans=map[x][i]+map[x-1][i-1];
}
printf("%d\n",ans+1);
}
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