HDU 1010 Tempter of the Bone 骨头的诱惑
2016-03-09 12:07
337 查看
Description
The doggie found a bone in an ancient maze, which fascinated him a lot. However, when hepicked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-thsecond for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block,the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.INPUT
The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), whichdenote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:'X': a block of wall, which the doggie cannot enter; 'S': the start point of the doggie; 'D': the Door; or'.': an empty block.The input is terminated with three 0's. This test case is not to be processed.OUTPUT
For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.SAMPLE INPUT
4 4 5 S.X. ..X. ..XD .... 3 4 5 S.X. ..X. ...D 0 0 0
SAMPLE OUTPUT
NO YES
一只小狗受骨头的诱惑进入了一个迷宫,初始时刻为0,小狗每走一步时间+1,迷宫的门只有在t时刻的时候打开一下,问小狗能不能走出迷宫(t时刻刚好到达迷宫的门)。‘S’表示小狗初始位置,‘D’表示门到的位置,‘.’表示空地(可行走),‘X’表示墙(不能行走)。每个位置只能行走一次。
这题还是用DFS做,但要转个弯。
刚开始直接做超时了三四次,然后去百度,学到了一招奇偶性剪枝,也就是我说的那个弯。
我这里简单说一下奇偶性剪枝。
地图上某点坐标为(x,y),如果x+y为奇数,这个点就是奇数点,x+y为偶数,这个点就是偶数点。
地图上奇偶点总是交替存在的,如下图,我用0表示偶数点,1表示奇数点:
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
如果小狗从偶数点出发,时间t就相当于它走的步数,如果走偶数步的话,只能到达偶数点。这时候如果门在奇数点,小狗就永远不能刚好在t时刻停在门。这时候就没必要进行DFS,进行了一次剪枝。在其他的奇偶位置也同理。
然后是代码:
#include<stdio.h>#include<string.h>#include<iostream>using namespace std;#define MAX 1005char mpt[MAX][MAX];int vis[MAX][MAX];int a,b,time,ans;//ans用于标记是否确定能出去int turn[4][2] = { 0, -1, 0, 1, 1, 0, -1,0};void DFS(int x,int y,int t){if (ans)return;if (t == 0){if (mpt[x][y] == 'D')//t剩余0,并且当前刚好在D上,ans=1,能出去{ans = 1;}return;//不管能不能出去,此时t已经等于0了,门只打开一下,继续走以及没用了,所以直接返回}int ty, tx;int i;for (i = 0; i < 4; i++){if (ans)return;tx = x+ turn[i][0];ty = y + turn[i][1];if (tx < 0 || ty < 0 || tx >= a || ty >= b)continue;if (vis[tx][ty] || mpt[tx][ty] == 'X')continue;vis[tx][ty] = 1;DFS(tx, ty, t - 1);vis[tx][ty] = 0;}}int main(){while (cin >> a>>b>>time){if (!a&&!b&&!time)return 0;int i,j,si,sj,di=-1,dj=-1;ans = 0;memset(mpt, 0, sizeof(mpt));memset(vis, 0, sizeof(vis));for (i = 0; i < a; i++){cin >> mpt[i];}for (i = 0; i < a; i++){for (j = 0; j < b; j++){if (mpt[i][j] == 'S'){si = i; sj = j;vis[si][sj] = 1;}else if (mpt[i][j] == 'D'){di = i; dj = j;}}}if (di == -1)ans = 0;//没有门,剪枝else if ((si + sj + time) % 2 != (di + dj) % 2)ans = 0;//**奇偶性剪枝else DFS(si, sj, time);if (ans)cout << "YES" << endl;else cout << "NO" << endl;}return 0;}
相关文章推荐
- Linux模块卸载
- 添加 MyEclipse Persistence Tools 类库
- C Primer Plus(九) 函数
- conetOS 笔记
- Android异步消息的处理机制(looper handler message)
- Netty学习之路(2)
- JSON.NET对象序列化示例教程
- 滚动单元格到某处
- 10.10.5升级到10.11后Cocoapods不可用的解决
- Lock
- [mybatis]动态SQL与SQL片段
- 延迟X秒之后执行某段代码
- Linux chmod命令修改文件与文件夹权限命令代码
- 经典排序算法(1) - 冒泡排序Bubble Sort
- 最全面的I/O优化介绍 从文件系统到磁盘管理
- 常用正则表达式
- [LeetCode]53. Maximum Subarray
- 140.Examine the structure of the PRODUCTS table:
- 字典树模板题&hdu1251
- 【python】编程语言入门经典100例--29