博客
关于我
POJ 2312:Battle City(BFS)
阅读量:217 次
发布时间:2019-02-28

本文共 3114 字,大约阅读时间需要 10 分钟。

                                            Battle City

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 9885   Accepted: 3285

Description

Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now. 

What we are discussing is a simple edition of this game. Given a map that consists of empty spaces, rivers, steel walls and brick walls only. Your task is to get a bonus as soon as possible suppose that no enemies will disturb you (See the following picture). 

Your tank can't move through rivers or walls, but it can destroy brick walls by shooting. A brick wall will be turned into empty spaces when you hit it, however, if your shot hit a steel wall, there will be no damage to the wall. In each of your turns, you can choose to move to a neighboring (4 directions, not 8) empty space, or shoot in one of the four directions without a move. The shot will go ahead in that direction, until it go out of the map or hit a wall. If the shot hits a brick wall, the wall will disappear (i.e., in this turn). Well, given the description of a map, the positions of your tank and the target, how many turns will you take at least to arrive there?

Input

The input consists of several test cases. The first line of each test case contains two integers M and N (2 <= M, N <= 300). Each of the following M lines contains N uppercase letters, each of which is one of 'Y' (you), 'T' (target), 'S' (steel wall), 'B' (brick wall), 'R' (river) and 'E' (empty space). Both 'Y' and 'T' appear only once. A test case of M = N = 0 indicates the end of input, and should not be processed.

Output

For each test case, please output the turns you take at least in a separate line. If you can't arrive at the target, output "-1" instead.

Sample Input

3 4YBEBEERESSTE0 0

Sample Output

8

题意

n*m的矩阵,Y代表起点,T代表终点,R不能通过,走E需要一步,B需要两步。求从起点到终点的最短距离。如果不能到达,输出-1

AC代码

#include 
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define ll long long#define ms(a) memset(a,0,sizeof(a))#define pi acos(-1.0)#define INF 0x3f3f3f3fconst double E=exp(1);const int maxn=1e3+10;char ch[maxn][maxn];using namespace std;int place[5][2]={1,0,-1,0,0,1,0,-1};int vis[maxn][maxn];int n,m;struct node{ int x,y,dis;}; bool operator < (const node a,const node b){ return a.dis>b.dis;}void bfs(int a,int b,int c,int d){ ms(vis); vis[a][b]=1; priority_queue
que; node start,end; start.x=a; start.y=b; start.dis=0; que.push(start); int ans=-1; while(!que.empty()) { start=que.top(); que.pop(); if(start.x==c&&start.y==d) { ans=start.dis; break; } for(int i=0;i<4;i++) { end.x=start.x+place[i][0]; end.y=start.y+place[i][1]; if(ch[end.x][end.y]=='R'||ch[end.x][end.y]=='S') continue; if(end.x<0||end.x>=n||end.y<0||end.y>=m) continue; if(vis[end.x][end.y]) continue; if(ch[end.x][end.y]=='E'||ch[end.x][end.y]=='T') end.dis=start.dis+1; if(ch[end.x][end.y]=='B') end.dis=start.dis+2; que.push(end); vis[end.x][end.y]++; } } cout<
<
>n>>m) { if(n==0&&m==0) break; ms(vis); ms(ch); int x1,x2,y1,y2; for(int i=0;i
>ch[i]; for(int i=0;i

 

转载地址:http://dcbp.baihongyu.com/

你可能感兴趣的文章
mysql的配置文件参数
查看>>
MySQL的错误:No query specified
查看>>
mysql监控工具-PMM,让你更上一层楼(上)
查看>>
mysql监控工具-PMM,让你更上一层楼(下)
查看>>
MySQL相关命令
查看>>
mysql社工库搭建教程_社工库的搭建思路与代码实现
查看>>
Warning: Can't perform a React state update on an unmounted component. This is a no-
查看>>
mysql笔记 (早前的,很乱)
查看>>
MySQL笔记:InnoDB的锁机制
查看>>
mysql第一天~mysql基础【主要是DDL、DML、DQL语句,以及重点掌握存存引擎、查询(模糊查询)】
查看>>
mysql第二天~mysql基础【查询排序、分页查询、多表查询、数据备份与恢复等】
查看>>
MySQL简介和安装
查看>>
MySQL简单查询
查看>>
MySQL管理利器 MySQL Utilities 安装
查看>>
MySQL篇(管理工具)
查看>>
mysql类型转换函数convert与cast的用法
查看>>
mysql系列一
查看>>
MySQL系列之数据授权(安全)
查看>>
MySQL系列之数据类型(Date&Time)
查看>>
MySQL系列之数据类型(Date&Time)
查看>>