zoukankan      html  css  js  c++  java
  • 搜索(BFS)

    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 <= R, C<= 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.

    Sample Input

    2
    4 4
    ####
    #JF#
    #..#
    #..#
    3 3
    ###
    #J.
    #.F
    

    Output Specification

    For each test case, output a single line containing IMPOSSIBLE if Joe cannot exit the maze before the fire reaches him, or an integer giving the earliest time Joe can safely exit the maze, in minutes.

    Output for Sample Input

    3
    IMPOSSIBLE
    #include <iostream>
    #include <cstdlib>
    #include <cstring>
    #include <cstdio>
    #include <algorithm>
    #include <string>
    #include <queue>
    #include <vector>
    
    using namespace std;
    
    #define INF 1<<29
    #define eps 1e-8
    #define A system("pause")
    #define rep(i,h,n) for(int i=(h);i<=(n);i++)
    #define ms(a,b) memset((a),(b),sizeof(a))
    #define lson l,mid,rt<<1
    #define rson mid+1,r,rt<<1|1
    
    const int maxn=2000+5;
    int dx[]={0,1,-1,0};
    int dy[]={1,0,0,-1};
    struct dot
    {
        int x,y;
        bool isfire;
        int time;
    };
    
    int vis[maxn][maxn];
    char map[maxn][maxn];
    int test,n,m;
    
    inline bool in(dot sgx)
    {
         if(sgx.x>=0&&sgx.x<n&&sgx.y>=0&&sgx.y<m) return true;
         return false;
    }
    
    int main()
    {
        scanf("%d",&test);
        while(test--)
        {
             dot gx;
             scanf("%d%d",&n,&m);
             rep(i,0,n-1) scanf("%s",map[i]);
             queue<dot> q;
             while(!q.empty()) q.pop();
             ms(vis,0);
             rep(i,0,n-1) rep(j,0,m-1)
             {dot F;
                  if(map[i][j]=='J') gx.x=i,gx.y=j,gx.isfire=0,gx.time=0;
                  else if(map[i][j]=='F')   vis[i][j]=1,F.x=i,F.y=j,F.isfire=1,F.time=0,q.push(F);
                  else if(map[i][j]=='#')   vis[i][j]=1;
             }
             q.push(gx);
             int ret=0;
             while(!q.empty())
             {
                 dot next;
                 gx=q.front(),q.pop();
                 rep(i,0,3)
                 {
                     next.x=gx.x+dx[i];
                     next.y=gx.y+dy[i];
                     next.isfire=gx.isfire;//BFS的时候注意"火势蔓延",相邻格子火势是相通的;
                     next.time=gx.time+1;
                     if(in(next))
                     {
                         if(!vis[next.x][next.y]) vis[next.x][next.y]=1,q.push(next);
                     }
                     else if(next.isfire==0)
                     {
                         ret=next.time;//记录当前答案;
                         break;
                     }
                 }
                 if(ret>0)break;
             }
             if(!ret) std::puts("IMPOSSIBLE");
             else printf("%d
    ",ret);
        }
        //A;
        return 0;
    }
    
  • 相关阅读:
    一种线程安全的handle
    基于数组的无锁队列(译)
    distri.lua的web运维工具
    distri.lua重写开源手游服务器框架Survive
    99 Lisp Problems 二叉树(P54~P69)
    99 Lisp Problems 列表处理(P1~P28)
    TSPL学习笔记(4):数组相关练习
    TSPL学习笔记(3):排序算法练习
    用ECMAScript4 ( ActionScript3) 实现Unity的热更新 -- 热更新Live2D
    用ECMAScript4 ( ActionScript3) 实现Unity的热更新 -- 使用FairyGUI (二)
  • 原文地址:https://www.cnblogs.com/scottding/p/3712838.html
Copyright © 2011-2022 走看看