Game III

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 31 Accepted Submission(s) : 11

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

Zjt and Sara will take part in a game, named Game III. Zjt and Sara will be in a maze, and Zjt must find Sara. There are some strang rules in this maze. If Zjt move a step, Sara will move a step in opposite direction.
Now give you the map , you shold find out the minimum steps, Zjt have to move. We say Zjt meet Sara, if they are in the same position or they are adjacent .
Zjt can only move to a empty position int four diraction (up, left, right, down). At the same time, Sara will move to a position in opposite direction, if there is empty. Otherwise , she will not move to any position.
The map is a N*M two-dimensional array. The position Zjt stays now is marked Z, and the position, where Sara stays, is marked E.

>  . : empty position
>  X: the wall
>  Z: the position Zjt now stay
>  S: the position Sara now stay

Your task is to find out the minimum steps they meet each other.

Input

The input contains several test cases. Each test case starts with a line contains three number N ,M (2<= N <= 20, 2 <= M <= 20 ) indicate the size of the map. Then N lines follows, each line contains M character. A Z and a S will be in the map as the discription above.

Output

For each test case, you should print the minimum steps. “Bad Luck!” will be print, if they can't meet each other.

Sample Input

4 4
XXXX
.Z..
.XS.
XXXX
4 4
XXXX
.Z..
.X.S
XXXX
4 4
XXXX
.ZX.
.XS.
XXXX

Sample Output

1
1
Bad Luck!

#include <iostream>
#include <cstdio>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
struct node
{int x,y,u,v,ti;node(int a,int b,int c,int d,int e){x=a;y=b;u=c;v=d;ti=e;}
};
int n,m,i,j,sx,sy,tx,ty,ans;
char mp[25][25];
int vis[25][25][25][25];  //标记走过否
int dr[4][2]={{0,1},{0,-1},{-1,0},{1,0} };bool check(int x,int y)
{if (x>=0 && x<n && y>=0 && y<m && mp[x][y]!='X') return 1;return 0;
}
void bfs()
{queue<node>Q;vis[sx][sy][tx][ty]=1;Q.push(node(sx,sy,tx,ty,0));while(!Q.empty()){node p=Q.front();Q.pop();if (abs(p.x-p.u)+abs(p.y-p.v)<=1) { ans=p.ti; return; }for(i=0;i<4;i++){int xx=p.x+dr[i][0];int yy=p.y+dr[i][1];int uu=p.u-dr[i][0];int vv=p.v-dr[i][1];if (check(xx,yy)){if (!check(uu,vv)){uu=p.u;vv=p.v;}if (!vis[xx][yy][uu][vv]){vis[xx][yy][uu][vv]=1;Q.push(node(xx,yy,uu,vv,p.ti+1));}}}}return;
}
int main()
{while(~scanf("%d%d",&n,&m)){for(i=0;i<n;i++){scanf("%s",&mp[i]);for(j=0;j<m;j++){if (mp[i][j]=='Z') sx=i,sy=j;if (mp[i][j]=='S') tx=i,ty=j;}}ans=-1;memset(vis,0,sizeof(vis));bfs();if (ans==-1) printf("Bad Luck!\n");else printf("%d\n",ans);}return 0;
}

转载于:https://www.cnblogs.com/stepping/p/5669304.html

HDU2216:Game III(BFS)相关推荐

leetcode 1306. Jump Game III | 1306. 跳跃游戏 III（BFS）
题目 https://leetcode.com/problems/jump-game-iii/ 题解正如 hint 所说: 用 BFS,跳过的就不跳了,直到全部跳过,即 count == arr.l ...
leetcode1306. 跳跃游戏 III（bfs）
这里有一个非负整数数组 arr,你最开始位于该数组的起始下标 start 处.当你位于下标 i 处时,你可以跳到 i + arr[i] 或者 i - arr[i]. 请你判断自己是否能够跳到对应元素值 ...
LeetCode 248. 中心对称数 III（DFS/BFS）
文章目录 1. 题目 2. 解题 2.1 DFS 2.2 BFS 1. 题目中心对称数是指一个数字在旋转了 180 度之后看起来依旧相同的数字(或者上下颠倒地看). 写一个函数来计算范围在 [low ...
LeetCode 1306. 跳跃游戏 III（广度优先搜索BFS）
1. 题目这里有一个非负整数数组 arr,你最开始位于该数组的起始下标 start 处. 当你位于下标 i 处时,你可以跳到 i + arr[i] 或者 i - arr[i]. 请你判断自己是否能够 ...
bzoj4997 [Usaco2017 Feb]Why Did the Cow Cross the Road III（bfs）
枚举每一个起点,bfs能访问到几头牛即可.复杂度O(n3)O(n^3) #include <cstdio> #include <cstring> #include <al ...
牛客编程巅峰赛：Tree III（BFS or DFS）
链接:https://ac.nowcoder.com/acm/contest/9557/C 来源:牛客网题目描述给出一棵有n个节点的节点标号为1~n的有根树(根为第一个节点,并给出从第2个节点到第 ...
剑指Offer - 面试题32 - III. 从上到下打印二叉树 III（BFS，queue+stack）
1. 题目请实现一个函数按照之字形顺序打印二叉树, 即第一行按照从左到右的顺序打印, 第二层按照从右到左的顺序打印, 第三行再按照从左到右的顺序打印,其他行以此类推. 例如: 给定二叉树: [3,9 ...
剑指offer面试题32 - III. 从上到下打印二叉树 III（二叉树)(BFS)
题目描述请实现一个函数按照之字形顺序打印二叉树,即第一行按照从左到右的顺序打印,第二层按照从右到左的顺序打印,第三行再按照从左到右的顺序打印,其他行以此类推. 思路详见链接代码 class So ...
【学时总结】 ◆学时·III◆ 二分图
[学时·III] 二分图 ■基本策略■ 其实本质是图论中的网络流二分图是两个由多个点组成的集合(上部和下部,且没有重叠),两个集合中的点不与该集合内其他的点连通,但和另一个集合内的点连通.我们称这两 ...

HDU2216:Game III(BFS)