poj 3182 The Grove

The Grove

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 641		Accepted: 297

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

思路：

突破口肯定是必须要围绕果园走一圈了，那么起点到一个点分顺时针和逆时针经过果园用bfs求两次最短路就够了，比赛时想了很久都没想到处理顺时针和逆时针的方法，就只能想到这步了，思维还是不能突破呀。

怎样处理顺时针和逆时针呢？从果园中引一条射线出去与地图边界相交，从起点出发两次bfs，一次控制只能向上经过射线，一次控制只能向下进过射线就够了。因为绕果园必须要经过射线上一点，所以最后用射线上的点来更新答案就OK了。

注意：

那个射线不能随便作的，要保证射线不与果园再次相交。想一想，为什么？

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
//#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 105
#define mod 1000000007
#define INF 0x3f3f3f3f
using namespace std;

typedef long long ll;
int n,m,ans;
int sx,sy;
int dx[]={-1,1,0,0,-1,-1,1,1};
int dy[]={0,0,-1,1,-1,1,-1,1};
int dist[maxn][maxn][2];
bool vis[maxn][maxn][2];
char mp[maxn][maxn];
char s[maxn];
struct Node
{
    int x,y;
}cur,now;
queue<Node>q;

void presolve()
{
    int i,j,k,flag=0;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        {
            if(mp[i][j]=='X')
            {
                flag=1;
                for(k=j-1;k>=1;k--)  // 将射线标记
                {
                    if(mp[i][k]=='X') continue ;
                    mp[i][k]='Y';
                }
                break ;
            }
        }
        if(flag) break ;
    }
}
void bfs(int k)
{
    int i,j,nx,ny,tx,ty;
    while(!q.empty()) q.pop();
    cur.x=sx;
    cur.y=sy;
    dist[sx][sy][k]=0;
    vis[sx][sy][k]=1;
    q.push(cur);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        nx=now.x;
        ny=now.y;
        for(i=0;i<8;i++)
        {
            tx=nx+dx[i];
            ty=ny+dy[i];
            if(tx<1||tx>n||ty<1||ty>m||mp[tx][ty]=='X'||vis[tx][ty][k]) continue ;
            if(k&&mp[tx][ty]=='Y')           // 到Y时控制过去的方向就好了
            {
                if(i==1||i==6||i==7) continue ;
            }
            else if(!k&&mp[tx][ty]=='Y')     // 到Y时控制过去的方向就好了
            {
                if(i==0||i==4||i==5) continue ;
            }
            cur.x=tx;
            cur.y=ty;
            dist[tx][ty][k]=dist[nx][ny][k]+1;
            vis[tx][ty][k]=1;
            q.push(cur);
        }
    }
}
void solve()
{
    int i,j;
    ans=INF;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        {
            if(mp[i][j]=='Y')
            {
                ans=min(ans,dist[i][j][0]+dist[i][j][1]);
            }
        }
    }
}
int main()
{
    int i,j,t;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%s",s);
            for(j=1;j<=m;j++)
            {
                mp[i][j]=s[j-1];
                if(mp[i][j]=='*') sx=i,sy=j;
            }
        }
        presolve();
        memset(vis,0,sizeof(vis));
        bfs(0);
        bfs(1);
        solve();
        printf("%d
",ans);
    }
    return 0;
}