广搜BFS 做题记录(python)

POJ百练 4116 拯救行动

【有序队列】

1.此题存在"慢路"（有守卫的位置）与"快路"，遇到"慢路"需要消耗多倍时间/资源。故需要给类加一个是否"可直接判断"累减属性，对于"慢路"，每次访问都减一，直到该属性转变为"可判断"，此后与正常通路点一致；

2.关于python题解版本，参考hzw大佬的代码，发现POJ上使用一维无限延伸列表+维护头尾指针构造的队列可以过（耗时不到4000ms），但使用collections库自带的双向队列一定会超时（>10000ms）。不知是什么原因，可能和评测程序环境、调用库函数的时间成本有关。总之，之后在题目不做要求时可以多考虑直接用列表构造队列。

参考代码1（hzw大佬）：

a = [''] * 200
xx = [0, 0, 1, -1]
yy = [-1, 1, 0, 0]
qx = [0] * 80000
qy = [0] * 80000
g = [0] * 80000
dis = [0] * 80000
mp = []

def bfs(bx, by):
    head = 0
    tail = 1
    qx[0] = bx
    qy[0] = by
    while head != tail:            
        x, y, st, d = qx[head], qy[head], g[head], dis[head]
        head = head + 1
        if st == 1:
            qx[tail] = x
            qy[tail] = y
            dis[tail] = d + 1
            g[tail] = 0
            tail = tail + 1
            continue
        for i in range(4):
            tx = x + xx[i]
            ty = y + yy[i]
            if tx < 0 or ty < 0 or tx >= n or ty >= m or a[tx][ty] == '#':
                continue
            if mp[tx][ty] == 0:
                if a[tx][ty] == 'a':
                    return d + 1
                mp[tx][ty] = 1
                qx[tail] = tx
                qy[tail] = ty
                g[tail] = (a[tx][ty] == 'x')
                dis[tail] = d + 1
                tail = tail + 1
    return -1

t = int(input())
for test in range(t):
    mp.clear()
    n, m = map(int, input().split())
    for i in range(n):
        mp.append([0] * 200)
        a[i] = input()
        for j in range(m):
            if(a[i][j] == 'r'):
                bx, by = i, j
    ans = bfs(bx, by)
    if ans == -1:
        print('Impossible')
    else:
        print(ans)

参考代码2（在hzw大佬版本上、根据个人习惯的调整）：

maze = ["" for i in range(200)]
dirx = [0,0,1,-1]
diry = [-1,1,0,0]
xseq = [0 for i in range(80000)]
yseq = [0 for i in range(80000)]
gd = [0 for i in range(80000)]
stp = [0 for i in range(80000)]
vis = []
n = 0
m = 0
def bfs(sx,sy):
    global maze,dirx,diry,xseq,yseq,gd,stp,vis,n,m
    hp = 0
    tp = 1
    xseq[0],yseq[0] = sx,sy
    while hp != tp:
        x,y,block,step = xseq[hp], yseq[hp], gd[hp], stp[hp]
        hp += 1
        if block == 1:
            xseq[tp] = x
            yseq[tp] = y
            stp[tp]  = step + 1
            gd[tp] = 0
            tp += 1
            continue
        for i in range(4):
            tx = x + dirx[i]
            ty = y + diry[i]
            if (tx<0 or ty<0 or tx>=n or ty>=m \
                or vis[tx][ty] == 1 or maze[tx][ty] == "#"):
                continue
            if maze[tx][ty] == "a":
                print(step+1)
                return
            if maze[tx][ty] == "x":
                gd[tp] = 1
            else:
                gd[tp] = 0
            xseq[tp],yseq[tp] = tx,ty
            vis[tx][ty] = 1
            stp[tp] = step+1
            tp += 1       
    print("Impossible")
    return
def solu():
    global maze,vis,n,m
    k = int(input())
    for i in range(k):
        vis.clear()
        rget = False
        n,m = map(int, input().split())
        for j in range(n):
            vis.append([0 for i in range(200)])
            maze[j] = input()
            if rget:
                continue
            for r in range(m):
                  if maze[j][r] == "r":
                      sx,sy = j,r
                      rget = True
        bfs(sx,sy)
solu()

参考代码3（deque库版本，POJ超时，仅供思路参考）：

import collections
class pt():
    def __init__(self,x_,y_,step_,gd_):
        self.x = x_
        self.y = y_
        self.step = step_
        self.gd = gd_

maze = ["" for i in range(205)]
dx = [0,0,1,-1]
dy = [1,-1,0,0]
vis = []
n,m = 0, 0

def bfs(sx,sy):
    global maze,vis,n,m
    q = collections.deque()
    q.append(pt(sx,sy,0,0))
    while len(q) > 0:
        ptmp = q.popleft()
        if ptmp.gd == 1:
            q.append(pt(ptmp.x, ptmp.y, ptmp.step+1, 0))
            continue
        for j in range(4):
            tx = ptmp.x + dx[j]
            ty = ptmp.y + dy[j]
            if (tx<0 or ty<0 or tx>=n or ty>=m \
                or vis[tx][ty]==1 or maze[tx][ty]=="#"):
                continue
            if maze[tx][ty] == "a":
                print(ptmp.step+1)
                return
            if maze[tx][ty] == "x":
                q.append(pt(tx,ty,ptmp.step+1,1))
            else:
                q.append(pt(tx,ty,ptmp.step+1,0))
    print("Impossible")
    return

def solu():
    global maze,vis,n,m
    k = int(input())
    for kk in range(k):
        rget = False
        vis.clear()
        n,m = map(int,input().split())
        for i in range(n):
            maze[i] = input()
            vis.append([0 for i in range(205)])
            if not rget:
                for j in range(m):
                    if maze[i][j] == "r":
                        sx_,sy_ = i,j
                        rget = True
        bfs(sx_,sy_)
    return
solu()

继续加油！