博客
关于我
Oh, my goddess(bfs)
阅读量:625 次
发布时间:2019-03-13

本文共 2628 字,大约阅读时间需要 8 分钟。

 

Oh, my goddess

时间限制:
3000 ms  |  内存限制:65535 KB
难度:
3
 
描述

Shining Knight is the embodiment of justice and he has a very sharp sword can even cleavewall. Many bad guys are dead on his sword.

One day, two evil sorcerer cgangee and Jackchess decided to give him some colorto see. So they kidnapped Shining Knight's beloved girl--Miss Ice! They built a M x Nmaze with magic and shut her up in it.

Shining Knight arrives at the maze entrance immediately. He can reach any adjacent emptysquare of four directions -- up, down, left, and right in 1 second. Or cleave one adjacent wall in 3

seconds, namely,turn it into empty square. It's the time to save his goddess! Notice: ShiningKnight won't leave the maze before he find Miss Ice.

 
输入
The input consists of blocks of lines. There is a blank line between two blocks.
The first line of each block contains two positive integers M <= 50 and N <= 50separated by one space. In each of the next M lines there is a string of length N contentsO and #.
O represents empty squares. # means a wall.
At last, the location of Miss Ice, ( x, y ). 1 <= x <= M, 1 <= y <= N.
(Shining Knight always starts at coordinate ( 1, 1 ). Both Shining and Ice's locationguarantee not to be a wall.)
输出
The least amount of time Shining Knight takes to save hisgoddess in one line.
样例输入
3 5O##########O#O#3 4
样例输出
14
题解:坑,水,搞定#花费3,过去1,总共应该是4;
代码:
1 #include
2 #include
3 #include
4 using namespace std; 5 struct Node{ 6 int x,y,step; 7 friend bool operator < (Node a,Node b){ 8 return a.step>b.step; 9 }10 };11 char map[55][55];12 int vis[55][55];13 int disx[4]={0,0,1,-1};14 int disy[4]={1,-1,0,0};15 int M,N,ex,ey;16 void bfs(int sx,int sy){17 memset(vis,0,sizeof(vis));18 vis[sx][sy]=1;19 priority_queue
dl;20 Node a,b;21 a.x=sx;a.y=sy;a.step=0;22 dl.push(a);23 while(!dl.empty()){24 a=dl.top();25 dl.pop();26 for(int i=0;i<4;i++){27 b.x=a.x+disx[i];b.y=a.y+disy[i];28 if(b.x<1||b.y<1||b.x>M||b.y>N||vis[b.x][b.y])continue;29 if(map[b.x][b.y]=='O'){30 b.step=a.step+1;31 }32 if(map[b.x][b.y]=='#')b.step=a.step+4;33 if(b.x==ex&&b.y==ey){34 printf("%d\n",b.step);35 return;36 }37 vis[b.x][b.y]=1;38 dl.push(b);39 }40 }41 }42 int main(){43 while(~scanf("%d%d",&M,&N)){44 for(int i=1;i<=M;i++)scanf("%s",map[i]+1);45 scanf("%d%d",&ex,&ey);46 bfs(1,1);47 }48 return 0;49 }

 

转载地址:http://izeaz.baihongyu.com/

你可能感兴趣的文章
MySQL Join算法与调优白皮书(二)
查看>>
Mysql order by与limit混用陷阱
查看>>
Mysql order by与limit混用陷阱
查看>>
mysql order by多个字段排序
查看>>
MySQL Order By实现原理分析和Filesort优化
查看>>
mysql problems
查看>>
mysql replace first,MySQL中处理各种重复的一些方法
查看>>
MySQL replace函数替换字符串语句的用法(mysql字符串替换)
查看>>
mysql replace用法
查看>>
Mysql Row_Format 参数讲解
查看>>
mysql select, from ,join ,on ,where groupby,having ,order by limit的执行顺序和书写顺序
查看>>
MySQL Server 5.5安装记录
查看>>
mysql server has gone away
查看>>
mysql skip-grant-tables_MySQL root用户忘记密码怎么办?修改密码方法:skip-grant-tables
查看>>
mysql slave 停了_slave 停止。求解决方法
查看>>
MySQL SQL 优化指南:主键、ORDER BY、GROUP BY 和 UPDATE 优化详解
查看>>
MYSQL sql语句针对数据记录时间范围查询的效率对比
查看>>
mysql sum 没返回,如果没有找到任何值,我如何在MySQL中获得SUM函数以返回'0'?
查看>>
mysql sysbench测试安装及命令
查看>>
mysql Timestamp时间隔了8小时
查看>>