POJ3182 The Grove[射线法+分层图最短路]

The Grove

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 904		Accepted: 444

Description

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.

Input

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).

Output

Line 1: The single line contains a single integer which is the smallest number of steps required to circumnavigate the grove.

Sample Input

6 7
.......
...X...
..XXX..
...XXX.
...X...
......*

Sample Output

13

Source

USACO 2006 January Silver

题意：

    一个n*m(n,m<=50)的矩阵有一片连着的树林，Bessie要从起始位置出发绕林子一圈再回来，每次只能向横着、竖着或斜着走一步。
问最少需多少步才能完成。

分析：

    1.如果搜出的路径能够包围其中的一个点，那么就能包围森林，这样问题就被简化了
    2.我们任选一个点作为被包围点，在搜索时利用射线法判断某这个点是否在多边形中
    3.判断点在多边形内外最常用的方法就是射线法，即以一条射线穿过多边形次数的奇偶性来判断。
　　　奇在偶不在。

//dp[x][y][0]表示从起点达到(x,y)的距离
//dp[x][y][1]表示从(x,y)到起点的距离+去的距离

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N=105;
const int dx[8]={0,0,1,-1,1,1,-1,-1};
const int dy[8]={1,-1,0,0,-1,1,-1,1};
int dp[N][N][2];char mp[N][N];
int n,m,sx,sy,gx,gy,px,py,pk,nx,ny,nk;
struct node{
    int x,y,k;
    node(int x=0,int y=0,int k=0):x(x),y(y),k(k){}
};
bool ok(){
    if(nx==gx&&ny<gy){
        if(px<nx) return 1;
    }
    if(px==gx&&py<gy){
        if(px>nx) return 1;
    }
    return 0;
}
void bfs(){
    memset(dp,-1,sizeof dp);
    queue<node>q;
    q.push(node(sx,sy,0));
    dp[sx][sy][0]=0;
    while(!q.empty()){
        node t=q.front();q.pop();
        px=t.x;py=t.y;pk=t.k;
        for(int i=0;i<8;i++){
            nx=px+dx[i];ny=py+dy[i];nk=pk;
            if(nx<1||ny<1||nx>n||ny>m||mp[nx][ny]=='X') continue;
            if(ok()) nk^=1;
            if(!(~dp[nx][ny][nk])){
                dp[nx][ny][nk]=dp[px][py][pk]+1;
                q.push(node(nx,ny,nk));
            }
        }
    }
    printf("%d\n",dp[sx][sy][1]);
}
int main(){
    scanf("%d%d",&n,&m);bool flag=0;
    for(int i=1;i<=n;i++) scanf("%s",mp[i]+1);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(mp[i][j]=='*') sx=i,sy=j;
            if(!flag&&mp[i][j]=='X') flag=1,gx=i,gy=j;
        }
    }
    bfs();
    return 0;
}