(BFS)1097: Yuchang and Zixiang ‘s maze

1097: Yuchang and Zixiang ‘s maze
Time Limit: 2 Sec Memory Limit: 128 MBSubmit: 863 Solved: 149
 
Description
 
One day , Yuchang and Zixiang go out of school to find some thing interesting . But both of them is LuChi , so the miss the way easily .
“Where am I ? who am I ?” Yuchang says . “Who attack you? You must want to say .” Zixiang adds . “Don’t say that , how we get out there ? I want my mom 555555……” Yuchang started crying . Zixiang
 become very panic . He doesn’t know how to get out there.
 Now , they find they are in a N*M maze , they are at the (a,b) point , they know they want to go to the point (c,d) , they want to finish as soon as possible, so , could you help them ?
Same as other maze , there are some point has boom , means they can’t get the point . Give you N,M and Num (the number of points that have booms . ) , then , Num lines contains pairs (x,y) means
 point (x,y) have booms . then , one line contains a , b , c , d ,the begin and end point .
They can mov forward , back , left and right . And every move cost 1 second . Calculate how many seconds they need to get to the finish point .
 
Input
 
The first line contains tow numbers N,M (0 < x,y < 1000)means the size of the maze.
The second line contains a number Num (0 < N < X*Y), means the number of points which have booms .
Then next N lines each contain two numbers , xi,yi , means (xi,yi) has a boom .
 
Output
One line , contains one number , the time they cost .
    If they can’t get to the finish point , output -1 .
 
Sample Input
1000 1000 4
5 5
5 7
4 6
6 6 
1 1 5 6
 
Sample Output
-1
 

题目大意:
 
余昌和子巷的迷宫
时间限制:2秒 内存限制:128mb submit: 863 已解决:149
 
描述
一天,余昌和子巷走出学校,发现了一些有趣的事情。但他们两人都是陆驰,所以很容易错过的方式。
“我在哪儿?”我是谁?”“谁攻击你?你一定想说。”子巷补充道。“别说了,我们怎么出去的?”我要我妈妈555555……”余昌哭了起来。子巷变得非常恐慌。他不知道怎么出去。
现在,他们发现他们在一个N*M的迷宫中,他们在(a,b)点,他们知道他们想去(c,d)点,他们想尽快完成,所以,你能帮助他们吗?
和其他迷宫一样,有一些点有boom,意思是他们不能得到那个点。给出N M和Num(有障碍的点的数量),则Num行包含对(x,y)表示点(x,y)有障碍。然后,一行包含a、b、c、d、起点和终点。它们可以向前、向后、向左和向右移动。每一步花费1秒。计算他们到达终点需要多少秒。
 
输入
第一行包含两个数字N,M (0 < x,y < 1000)表示迷宫的大小。
第二行包含一个数字Num (0 < N < X*Y),表示具有障碍的点的数量。
然后接下来N行每一行包含两个数字,xi,yi ,表示(xi,yi) 有一个障碍。
 
输出
一行,包含一个数字,它们花费的时间。
如果不能到达终点,输出-1。
 
样例输入
1000 1000 4
5 5
5 7
4 6
6 6 
1 1 5 6
 
样例输出
-1
 
分析
典型的走迷宫问题
小地图(200*200一下)BFS、DFS都可以。大地图的话,因为信息量很大所以用不了DFS会妥妥地T掉,只能用BFS。

 

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 
  4 const int maxn = 1010;
  5 char maze[maxn][maxn];    //存储迷宫
  6 int vis[maxn][maxn];    //存储是否访问过标记
  7 int step[maxn][maxn];    //存储步数
  8 
  9 int n, m;    //m,n分别是迷宫的大小,在check中判断越界要使用,需要声明为全局变量
 10 
 11 int Move[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};    //四个方向可走,使用for循环来对当前坐标进行上下左右移动
 12 
 13 struct point    //结构体存储x,y的坐标,结构体的对象放在队列中
 14 {
 15     int x, y;
 16 } in, out, beg;            
 17 
 18 int check(int x, int y)
 19 {
 20     if(vis[x][y] == 0 && x >= 1 && x <= n && y >= 1 && y <= m && maze[x][y] != '#')
 21         return 1;
 22     else
 23         return 0;
 24 }
 25 
 26 int bfs()
 27 {
 28     memset(vis, 0, sizeof(vis));    //初始所有坐标的访问设置为0
 29     memset(step, 0, sizeof(step));    //初始所有步数为0
 30     
 31     vis[beg.x][beg.y] = 1;    //beg坐标点开始标记为1,表示已经访问
 32     step[beg.x][beg.y] = 0;    //beg所在的坐标为第一个,步数为0
 33     
 34     queue<point>q;
 35     
 36     q.push(beg);    //开始的那个点进队
 37 
 38     while(!q.empty())
 39     {
 40         out = q.front();    //out记住队头元素
 41         q.pop();
 42         for(int i = 0; i < 4; i++)
 43         {
 44             in.x = out.x + Move[i][0];    //循环四次,in分别是队头元素的上、下、左、右邻接坐标点
 45             in.y = out.y + Move[i][1];
 46             if(check(in.x, in.y))    //对in作是否访问过、越界、是否是障碍检查
 47             {
 48                 if(maze[in.x][in.y] == 'E')    //判断in是否是到达的坐标
 49                 {
 50                     return step[out.x][out.y] + 1;    //若到达,返回in前一个坐标的步数+1
 51                 }
 52                 q.push(in);   //不是终点,继续将in进队   
 53                 vis[in.x][in.y] = 1;
 54                 step[in.x][in.y] = step[out.x][out.y] + 1;    //    将in的步数在它前一个坐标的步数+1
 55             }
 56         }
 57     }
 58     return -1;
 59 }
 60 
 61 int main()
 62 {
 63     while(~scanf("%d%d", &n, &m))
 64     {
 65         for(int i = 1; i <= n; i++)
 66         {
 67             for(int j = 1; j <= m; j++)
 68             {
 69                 maze[i][j] = '.';
 70             }
 71         }
 72         int t;
 73         scanf("%d", &t);
 74         while(t--)
 75         {
 76             int x, y;
 77             scanf("%d%d", &x, &y);
 78             maze[x][y] = '#';    //将障碍的坐标设置为#
 79         }
 80         int a, b, c, d;
 81         scanf("%d%d%d%d", &a, &b, &c, &d);
 82         if(a == c && b == d)    //判断如果出发点和终点坐标一样,就直接输出0
 83         {
 84             printf("0\n");
 85         }
 86         else    //否则进行广度搜索
 87         {
 88             maze[a][b] = 'S';
 89             maze[c][d] = 'E';
 90             for(int i = 1; i <= n; i++)
 91             {
 92                 for(int j = 1; j <= m; j++)
 93                 {
 94                     if(maze[i][j] == 'S')    //    找到起始点,将起始点的坐标存入结构体变量beg中
 95                     {
 96                         beg.x = i;
 97                         beg.y = j;
 98                     }
 99                 }
100             }
101             int ans = bfs();
102             printf("%d\n", ans);
103         }
104     }
105     return 0;
106 }

 

 
posted @ 2019-03-14 11:44  WindSun  阅读(195)  评论(0编辑  收藏  举报
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