bzoj 1656: [Usaco2006 Jan] The Grove 树木（BFS）

1656: [Usaco2006 Jan] The Grove 树木

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 246 Solved: 158
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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.

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．

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．

Sample Input

6 7

.......

...X...

..XXX..

...XXX.

...X...

......*

Sample Output

题目保证所有的树木一定在一个联通块中，并且起始点一定不会被树包围

（也就是说至少有两个方向上没有树木）

那么步骤：

①输入时的第一个'X'，如果它上面没有'*'，那么将这个'X'上面所有的'.'全部置为'T'，否则就找下一个'X'

②从'*'处开始BFS，将'T'和'X'全部视为墙

③暴力所有的'T'，答案就是每个'T'两测点最短路和的最小值+2（注意斜着的也算）

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
typedef struct
{int x;int y;
}Point;
Point now, temp;
queue<Point> q;
char str[55][55];
int bet[55][55], dir[8][2] = {1,0,0,1,-1,0,0,-1,1,1,1,-1,-1,-1,-1,1};
int main(void)
{int n, m, i, j, x, y, k, fx, fy, l, r, ans;scanf("%d%d", &n, &m);memset(str, 'X', sizeof(str));x = y = fx = fy = -1;for(i=1;i<=n;i++){for(j=1;j<=m;j++){scanf(" %c", &str[i][j]);if(str[i][j]=='X' && fx==-1 && (y!=j || x==-1 || x>=i)){for(k=1;k<=i-1;k++)str[k][j] = 'T';fx = i, fy = j;}if(str[i][j]=='*')x = i, y = j;}}now.x = x, now.y = y;memset(bet, 62, sizeof(bet));q.push(now);bet[now.x][now.y] = 0;while(q.empty()==0){now = q.front();q.pop();for(k=0;k<=7;k++){temp.x = now.x+dir[k][0];temp.y = now.y+dir[k][1];if(str[temp.x][temp.y]!='X' && str[temp.x][temp.y]!='T' && bet[now.x][now.y]+1<bet[temp.x][temp.y]){bet[temp.x][temp.y] = bet[now.x][now.y]+1;q.push(temp);}}}ans = 100000;for(i=1;i<=fx-1;i++){j = fy;l = min(bet[i][j-1], min(bet[i-1][j-1], bet[i+1][j-1]));r = min(bet[i][j+1], min(bet[i-1][j+1], bet[i+1][j+1]));ans = min(ans, l+r+2);}printf("%d\n", ans);return 0;
}