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. 趋势型指标——MACD

    1.简要介绍▪ 计算方法▪ DIFF▪ DEA▪ MACD▪ 构造原理▪ 缺点2.实战技巧3.运用技巧▪ 应用原理▪ 经典用法▪ 实战战法▪ 捕捉卖点▪ 买卖策略▪ 短线实战4.组合指标运用5.一般研 ...

  2. [jQuery学习系列一]1-选择器与DOM对象

    前言: 好久没有更新博客了, 最近想复习下 之前学过的JS的相关内容, 也算是自己的一种总结. 知识长时间不用就会忘记, 多学多记多用!! 下面的程序都可以在下面的网站进行在线调试: http://w ...

  3. linux学习笔记1

    2016年09月25日 unix与linux发展史 unix是linux之父,学习linux后能容易上手unix. linux是源码开放的unix,由一位芬兰大学生李纳斯在网络上发起,和后来众多爱好者 ...

  4. TF Boys (TensorFlow Boys ) 养成记(六)

    圣诞节玩的有点嗨,差点忘记更新.祝大家昨天圣诞节快乐,再过几天元旦节快乐. 来继续学习,在/home/your_name/TensorFlow/cifar10/ 下新建文件夹cifar10_train ...

  5. 匿名管道读取CMD回显信息

    之前用了很坑爹的做法去读取了cmd命令的回显信息,现在发现了用匿名管道的实现方法,由于楼主没有学过Windows核心编程,找了一个代码来凑数 存下来以后研究 #include <windows. ...

  6. javaweb回顾第五篇浅谈会话

    1:什么是会话 通俗来说就是客户和服务器的一次私密谈话,客户发送请求以后服务器能够识别请求是来自同一个客户,他们是1对1的关系. 了解会话以后我们就要去考虑如何去实现这些问题下面一一进行解析 2:会话 ...

  7. GO語言視頻教程

    第1课:https://github.com/Unknwon/go-fundamental-programming/blob/master/lectures/lecture1.md Go开发环境搭建h ...

  8. JavaScript中如何中断forEach循环

    先来看下forEach的实现 // Production steps of ECMA-262, Edition 5, 15.4.4.18// Reference: http://es5.github. ...

  9. Utopian Tree in Java

    The Utopian tree goes through 2 cycles of growth every year. The first growth cycle occurs during th ...

  10. 解决Visual Studio 2010/2012的RC4011 warnings

    如果在vc10/11工程的rc文件中有以下任意一行代码: #include <winuser.h> #include <richedit.h> 那么vc将会给出一对警告: C: ...