【BZOJ】1656:[Usaco2006 Jan]The Grove 树木(bfs+特殊的技巧)










#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr2(a, b, c) for1(i, 1, b) { for1(j, 1, c) cout << a[i][j] << '\t'; cout << endl; }
#define printarr1(a, b) for1(i, 1, b) cout << a[i] << '\t'; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=60, Q=N*N*10, dx[]={-1, 1, 0, 0, -1, 1, -1, 1}, dy[]={0, 0, -1, 1, -1, -1, 1, 1};
int a[N][N], line[N][N], f[2][N][N], n, m, front, tail, flag, x, y, X, Y;
struct dat { int x, y, f; }q[Q];

int main() {
	read(n); read(m);
	for1(i, 1, n) for1(j, 1, m) {
		char ch=getchar(); while(ch!='.'&&ch!='X'&&ch!='*') ch=getchar();
		if(ch=='X') a[i][j]=1, x=i, y=j;
		else if(ch=='*') X=i, Y=j;
	for1(i, 1, n) if(i+x<=n) line[i+x][y]=1; else break;
	q[tail].x=X, q[tail].y=Y, q[tail++].f=0;
	while(front!=tail) {
		dat &t=q[front++]; if(front==Q) front=0;
		x=t.x, y=t.y, flag=t.f;
		rep(i, 8) {
			int fx=dx[i]+x, fy=dy[i]+y;
			if(fx<0 || fy<0 || fx>n || fy>m || a[fx][fy]) continue;
			if((line[x][y] || line[fx][fy]) && fy<=y) continue; //当前在射线上或下一个点是射线上时(即从右向左),不能向左拓展
			if(line[fx][fy] && !f[1][fx][fy]) { //当这是从射线左边向射线过来时,更新上的点
				q[tail].x=fx, q[tail].y=fy, q[tail++].f=1; if(tail==Q) tail=0;
			else if(!f[flag][fx][fy]) { //继承饶了一圈后的距离向四周拓展
				q[tail].x=fx, q[tail].y=fy, q[tail++].f=flag; if(tail==Q) tail=0;
	return 0;






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.


* 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).



* 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

Sample Output

posted @ 2014-09-22 06:13  iwtwiioi  阅读(406)  评论(0编辑  收藏  举报