Fire!

Time Limit:1000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Description

 

Problem B: Fire!

Joe works in a maze. Unfortunately, portions of the maze have caught on fire, and the owner of the maze neglected to create a fire escape plan. Help Joe escape the maze.

Given Joe's location in the maze and which squares of the maze are on fire, you must determine whether Joe can exit the maze before the fire reaches him, and how fast he can do it.

Joe and the fire each move one square per minute, vertically or horizontally (not diagonally). The fire spreads all four directions from each square that is on fire. Joe may exit the maze from any square that borders the edge of the maze. Neither Joe nor the fire may enter a square that is occupied by a wall.

Input Specification

The first line of input contains a single integer, the number of test cases to follow. The first line of each test case contains the two integers R and C, separated by spaces, with 1 <= RC <= 1000. The following R lines of the test case each contain one row of the maze. Each of these lines contains exactly C characters, and each of these characters is one of:

  • #, a wall
  • ., a passable square
  • J, Joe's initial position in the maze, which is a passable square
  • F, a square that is on fire

There will be exactly one J in each test case.

 
广度优先搜索~
 
 #include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <utility>
#include <cstdio>
#include <cstring> using namespace std; struct Pos
{
int y, x, step;
Pos(int a, int b, int c) {y = a; x = b; step = c;}
};
int r, c, fi, fj, ji, jj;
char m[][];
queue<Pos> que;
int fx[], fy[], top, tail; const int d[][] = {{, }, {, -}, {, }, {-, }}; bool Judge(int i, int j)
{
return i > - && j > - && i < r && j < c && m[i][j] == '.';
} bool Judge2(int i, int j)
{
return i > - && j > - && i < r && j < c && m[i][j] != '#' && m[i][j] != 'F';
} bool isExit(int i, int j)
{
return i == || j == || i == r - || j == c - ;
} void fire_spread()
{
int y, x, yy, xx, end = tail;
for(; top < end; top++){
y = fy[top];
x = fx[top];
for(int i = ; i < ; i++){
yy = y + d[i][];
xx = x + d[i][];
if(Judge2(yy, xx)){
m[yy][xx] = 'F';
fx[tail] = xx;
fy[tail] = yy;
tail++;
}
}
}
} int main()
{
int t;
scanf("%d", &t);
while(t--){
bool ok = false;
top = tail = ;
while(!que.empty()) que.pop();
scanf("%d %d", &r, &c);
for(int i = ; i < r; i++){
scanf("%s", m[i]);
for(int j = ; m[i][j] != '\0'; j++){
if(m[i][j] == 'J') que.push(Pos(i, j, ));
else if(m[i][j] == 'F') fy[tail] = i, fx[tail] = j, tail++;
}
}
int pre = ;
fire_spread();
while(!que.empty()){
Pos cur = que.front();
que.pop();
if(isExit(cur.y, cur.x)){
pre = cur.step + ;
ok = true;
break;
}
if(cur.step > pre) {
fire_spread();
pre++;
}
for(int i = ; i < ; i++){
int ii = cur.y + d[i][];
int jj = cur.x + d[i][];
if(Judge(ii, jj)) {
m[ii][jj] = 'P';
que.push(Pos(ii, jj, cur.step + ));
}
}
}
if(!ok) puts("IMPOSSIBLE");
else printf("%d\n", pre);
}
return ;
}

Fire!(BFS)的更多相关文章

  1. UVA 11624 Fire! (bfs)

    算法指南白书 分别求一次人和火到达各个点的最短时间 #include<cstdio> #include<cstring> #include<queue> #incl ...

  2. UVA - 11624 Fire! bfs 地图与人一步一步先后搜/搜一次打表好了再搜一次

    UVA - 11624 题意:joe在一个迷宫里,迷宫的一些部分着火了,火势会向周围四个方向蔓延,joe可以向四个方向移动.火与人的速度都是1格/1秒,问j能否逃出迷宫,若能输出最小时间. 题解:先考 ...

  3. UVA11624 Fire! —— BFS

    题目链接:https://vjudge.net/problem/UVA-11624 题解: 坑点:“portions of the maze havecaught on fire”, 表明了起火点不唯 ...

  4. UVA 11624 Fire! BFS搜索

    题意:就是问你能不能在火烧到你之前,走出一个矩形区域,如果有,求出最短的时间 分析:两遍BFS,然后比较边界 #include<cstdio> #include<algorithm& ...

  5. UVA 11624 Fire! bfs 难度:0

    http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&p ...

  6. ACM: FZU 2150 Fire Game - DFS+BFS+枝剪 或者 纯BFS+枝剪

    FZU 2150 Fire Game Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u ...

  7. UVa 11624 Fire!(BFS)

    Fire! Time Limit: 5000MS   Memory Limit: 262144KB   64bit IO Format: %lld & %llu Description Joe ...

  8. foj 2150 Fire Game(bfs暴力)

         Problem Description Fat brother and Maze are playing a kind of special (hentai) game on an N*M ...

  9. fzu 2150 Fire Game 【身手BFS】

    称号:fzupid=2150"> 2150 Fire Game :给出一个m*n的图,'#'表示草坪,' . '表示空地,然后能够选择在随意的两个草坪格子点火.火每 1 s会向周围四个 ...

随机推荐

  1. Java程序员的日常——经验贴(纯干货)

    工作当中遇到的事情比较杂,因此涉及的知识点也很多.这里暂且记录一下,今天遇到的知识点,纯干货~ 关于文件的解压和压缩 如果你的系统不支持tar -z命令 如果是古老的Unix系统,可能并不认识tar ...

  2. 使用Python的yield实现流计算模式

    首先先提一下上一篇<如何猜出Y combinator>中用的方法太复杂了.其实在Lambda演算中实现递归的思想很简单,就是函数把自己作为第一个参数传入函数,然后后面就是简单的Lambda ...

  3. Maven学习总结(五)——聚合与继承

    一.聚合 如果我们想一次构建多个项目模块,那我们就需要对多个项目模块进行聚合 1.1.聚合配置代码 <modules> <module>模块一</module> & ...

  4. Multiplexing SDIO Devices Using MAX II or CoolRunner-II CPLD

    XAPP906 Supporting Multiple SD Devices with CoolRunner-II CPLDs There has been an increasing demand ...

  5. 无锁编程以及CAS

    无锁编程 / lock-free / 非阻塞同步 无锁编程,即不使用锁的情况下实现多线程之间的变量同步,也就是在没有线程被阻塞的情况下实现变量的同步,所以也叫非阻塞同步(Non-blocking Sy ...

  6. solr课程学习系列-solr服务器配置(2)

    本文是solr课程学习系列的第2个课程,对solr基础知识不是很了解的请查看solr课程学习系列-solr的概念与结构(1) 本文以windows的solr6服务器搭建为例. 一.solr的工作环境: ...

  7. SSD在SQLServer中的应用

        一. 首先,回顾一下 SSD 的读写特性 (1)有限次数写:        (2)随机读性能最好:        (3)顺序读性能好:        (4)顺序写性能差:        (5) ...

  8. 从一个例子中体会React的基本面

    [起初的准备工作] npm init npm install --save react react-dom npm install --save-dev html-webpack-plugin web ...

  9. Jmeter报告优化之New XSL stylesheet

    Jmeter默认的报告展示的信息比较少,如果出错了,不是很方便定位问题.由Jmeter默认报告优化这篇文章可知,其实由.jtl格式转换为.html格式的报告过程中,style文件起了很关键的作用.下面 ...

  10. [leetcode]Find Minimum in Rotated Sorted Array II @ Python

    原题地址:https://oj.leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/ 解题思路:这道题和上一道题的区别是,数组中 ...