# 【BZOJ-1656】The Grove 树木 BFS + 射线法

## 1656: [Usaco2006 Jan] The Grove 树木

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 186  Solved: 118
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## 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.

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．

## 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).

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．

## Output

* 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．

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

13

Silver

## Code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
int N,M,sx,sy,lx,ly,dis[2][51][51];
struct Pa{int x,y,d;};
int dx[9]={0,0,0,1,1,1,-1,-1,-1},dy[9]={0,1,-1,0,1,-1,0,1,-1};
bool visit[51][51],mp[51][51],line[51][51];
inline bool OK(int x,int y) {return x>=1 && x<=N && y>=1 && y<=M;}
void BFS()
{
queue<Pa>q; q.push((Pa){sx,sy,0}); dis[0][sx][sy]=1;
while (!q.empty())
{
Pa now=q.front(); q.pop();
for (int d=1,x,y; d<=8; d++)
{
x=now.x+dx[d],y=now.y+dy[d];
if (!OK(x,y) || mp[x][y]) continue;
if (line[now.x][now.y] && y<=now.y) continue;
if (line[x][y] && y>now.y)
if (!dis[1][x][y])
dis[1][x][y]=dis[0][now.x][now.y]+1,q.push((Pa){x,y,1});
else;
else
if (!dis[now.d][x][y])
dis[now.d][x][y]=dis[now.d][now.x][now.y]+1,q.push((Pa){x,y,now.d});
else;
}
}
}
int main()
{
scanf("%d%d",&N,&M); char c[51][51];
for (int i=1; i<=N; i++)
{
scanf("%s",c[i]+1);
for (int j=1; j<=M; j++)
mp[i][j]=c[i][j]=='X',lx=c[i][j]=='X'? i:lx,ly=c[i][j]=='X'? j:ly,
sx=c[i][j]=='*'? i:sx,sy=c[i][j]=='*'? j:sy;
}
for (int i=1; i<=N; i++) if (i+lx<=N) line[i+lx][ly]=1;
BFS();
printf("%d\n",dis[1][sx][sy]-1);
return 0;
}

posted @ 2016-10-26 21:22  DaD3zZ  阅读(579)  评论(0编辑  收藏  举报