Luogu P2864 [USACO06JAN]树林The Grove(bfs)

P2864 [USACO06JAN]树林The Grove(bfs)

题面

题目描述

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 ( \(1 \leq R \leq 50, 1 \leq C \leq 50\) ). 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( \(1 \leq R \leq 50，1 \leq C \leq 50\) ).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.

输入输出格式

输入格式：

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).

输出格式：

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

输入输出样例

输入样例：

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

输出样例：

13

思路

艹，这题的边界判断！ \(LSK\) 你试试这道题。 --zbtrs

换了三种写法，终于把神犇推荐的这道题 \(AC\) 了...

我们直接拿样例做例子：

.......
...X...
..XXX..
...XXX.
...X...
......*

比如说最上面的那一颗树，我们在它上面做一条射线。

.......
----
...X...
..XXX..
...XXX.
...X...
......*

那么绕树林一周的路线一定要经过这条射线。

我们可以用 \(bfs\) 更新所有点到出发点的最短距离，并且特判那条射线，要求不能跨过它：

typedef pair<int,int> PII;//个人代码习惯
#define mp(a,b) make_pair(a,b)//同上
int a[8]={-1,-1,-1,+0,+0,+1,+1,+1};//x的变化
int b[8]={-1,+0,+1,-1,+1,+1,+0,-1};//y的变化
step[sx][sy]=1;
queue<PII>Q;//队列
Q.push(mp(sx,sy));//放入出发点
while(!Q.empty())
{
    int x=Q.front().first,y=Q.front().second;Q.pop();//取出队首点
    for(int i=0;i<8;i++)//枚举所有情况
    {
        int dx=x+a[i],dy=y+b[i];//能到达的点
        if(step[dx][dy]||!G[dx][dy]) continue;//不能到达的点和已经到过的点
        if(y<=ly&&x==lx&&dx==lx-1) continue;//不能从上往下穿过射线
        if(y<=ly&&x==lx-1&&dx==lx) continue;//不能从下往上穿过射线
        step[dx][dy]=step[x][y]+1;
        Q.push(mp(dx,dy));
    }
}

然后就顺利 \(AC\) 了。

AC代码

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> PII;
#define mp(a,b) make_pair(a,b)
int n,m,sx,sy,lx,ly,ans=INT_MAX,G[55][55],step[55][55];
bool is_line_made;
int a[8]={-1,-1,-1,+0,+0,+1,+1,+1};
int b[8]={-1,+0,+1,-1,+1,+1,+0,-1};
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            char ch;
            cin>>ch;
            if(ch=='*') G[i][j]=1,sx=i,sy=j;
            else if(ch=='.') G[i][j]=1;
            else if(!lx) lx=i,ly=j;
        }
    step[sx][sy]=1;
    queue<PII>Q;
    Q.push(mp(sx,sy));
    while(!Q.empty())
    {
        int x=Q.front().first,y=Q.front().second;Q.pop();
        for(int i=0;i<8;i++)
        {
            int dx=x+a[i],dy=y+b[i];
            if(step[dx][dy]||!G[dx][dy]) continue;
            if(y<=ly&&x==lx&&dx==lx-1) continue;
            if(y<=ly&&x==lx-1&&dx==lx) continue;
            step[dx][dy]=step[x][y]+1;
            Q.push(mp(dx,dy));
        }
    }
    for(int i=1;i<=ly;i++)
    {
        if(step[lx][i]&&step[lx-1][i]&&ans>step[lx][i]+step[lx-1][i]) ans=step[lx][i]+step[lx-1][i];
        if(step[lx][i]&&step[lx-1][i+1]&&ans>step[lx][i]+step[lx-1][i+1]) ans=step[lx][i]+step[lx-1][i+1];
        if(step[lx][i]&&step[lx-1][i-1]&&ans>step[lx][i]+step[lx-1][i-1]) ans=step[lx][i]+step[lx-1][i-1];
    }
    printf("%d",ans-1);
    return 0;
}