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 网络目录_联机目录数据库
查看>>
MySQL 聚簇索引&&二级索引&&辅助索引
查看>>
Mysql 脏页 脏读 脏数据
查看>>
mysql 自增id和UUID做主键性能分析,及最优方案
查看>>
Mysql 自定义函数
查看>>
mysql 行转列 列转行
查看>>
Mysql 表分区
查看>>
mysql 表的操作
查看>>
mysql 视图,视图更新删除
查看>>
MySQL 触发器
查看>>
mysql 让所有IP访问数据库
查看>>
mysql 记录的增删改查
查看>>
MySQL 设置数据库的隔离级别
查看>>
MySQL 证明为什么用limit时,offset很大会影响性能
查看>>
Mysql 语句操作索引SQL语句
查看>>
MySQL 误操作后数据恢复(update,delete忘加where条件)
查看>>
MySQL 调优/优化的 101 个建议!
查看>>
mysql 转义字符用法_MySql 转义字符的使用说明
查看>>
mysql 输入密码秒退
查看>>
mysql 递归查找父节点_MySQL递归查询树状表的子节点、父节点具体实现
查看>>