题目地址: http://poj.org/problem?id=1475 两重BFS就行了,第一重是搜索箱子,第二重搜索人能不能到达推箱子的地方. AC代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <vector> #inc…

题目链接:http://poj.org/problem?id=1475 一组测试数据: 7 3 ### .T. .S. #B# ... ... ... 结果: //解题思路:先判断盒子的四周是不是有空位,如果有,则判断人是否能到达盒子的那一边,能的话,把盒子推到空位处,然后继续 AC代码: //解题思路:先判断盒子的四周是不是有空位,如果有,则判断人是否能到达盒子的那一边,能的话,把盒子推到空位处,然后继续 #include<iostream> #include<algorithm>…

Imagine you are standing inside a two-dimensional maze composed of square cells which may or may not be filled with rock. You can move north, south, east or west one cell at a step. These moves are called walks.  One of the empty cells contains a box…

[poj P1475] Pushing Boxes Time Limit: 2000MS   Memory Limit: 131072K   Special Judge Description Imagine you are standing inside a two-dimensional maze composed of square cells which may or may not be filled with rock. You can move north, south, east…

Pushing Boxes Time Limit: 2000ms Memory Limit: 131072KB This problem will be judged on PKU. Original ID: 147564-bit integer IO format: %lld      Java class name: Main Special Judge   Imagine you are standing inside a two-dimensional maze composed of…

http://poj.org/problem?id=1475 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=249 Pushing Boxes Time Limit: 2000MS   Memory Limit: 131072K Total Submissions: 4662   Accepted: 1608   Special Judge Description Imagine you are standing insid…

题目传送门 首先说明我这个代码和lyd的有点不同:可能更加复杂 既然要求以箱子步数为第一关键字,人的步数为第二关键字,那么我们可以想先找到箱子的最短路径.但单单找到箱子的最短路肯定不行啊,因为有时候不能被推动,怎样确定一条既满足最短又满足可行的箱子路径呢,其实这就是一种有限制的BFS. 对于箱子: 设现在的位置为x,y,扩展方向为dx,dy,将要到达的下一个位置为x+dx,y+dy check它是否可行: 1.不越界. 2.之前没有走过. 3.不能走到“#”上. 4.人能够从当前站立的位置到达(…

Pushing Boxes Description Imagine you are standing inside a two-dimensional maze composed of square cells which may or may not be filled with rock. You can move north, south, east or west one cell at a step. These moves are called walks. One of the e…

POJ1475 Pushing Boxes  推箱子,#表示墙,B表示箱子的起点,T表示箱子的目标位置,S表示人的起点 本题没有 Special Judge,多解时,先最小化箱子被推动的次数,再最小化人移动的步数.若仍有多条路线,则按照N.S.W.E的顺序优先选择箱子的移动方向(即先上下推,再左右推).在此前提下,再按照n.s.w.e的顺序优先选择人的移动方向(即先上下动,再左右动). 每个阶段的状态包括人的位置,箱子的位置,由于每次在移动箱子之后,认得位置一定位于箱子之前的位置,所以可以将每次…

[题目链接] http://poj.org/problem?id=1475 [算法] 双重BFS [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #include <cstdio> #incl…