HDUOJ-----(1072)Nightmare(bfs)
Nightmare
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5795 Accepted Submission(s): 2868
Problem Description
Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bomb is set to 6 minutes. To prevent the bomb from exploding by shake, Ignatius had to move slowly, that is to move from one area to the nearest area(that is, if Ignatius stands on (x,y) now, he could only on (x+1,y), (x-1,y), (x,y+1), or (x,y-1) in the next minute) takes him 1 minute. Some area in the labyrinth contains a Bomb-Reset-Equipment. They could reset the exploding time to 6 minutes. Given the layout of the labyrinth and Ignatius' start position, please tell Ignatius whether he could get out of the labyrinth, if he could, output the minimum time that he has to use to find the exit of the labyrinth, else output -1. Here are some rules: 1. We can assume the labyrinth is a 2 array. 2. Each minute, Ignatius could only get to one of the nearest area, and he should not walk out of the border, of course he could not walk on a wall, too. 3. If Ignatius get to the exit when the exploding time turns to 0, he can't get out of the labyrinth. 4. If Ignatius get to the area which contains Bomb-Rest-Equipment when the exploding time turns to 0, he can't use the equipment to reset the bomb. 5. A Bomb-Reset-Equipment can be used as many times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as many times as you wish. 6. The time to reset the exploding time can be ignore, in other words, if Ignatius get to an area which contain Bomb-Rest-Equipment, and the exploding time is larger than 0, the exploding time would be reset to 6.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case starts with two integers N and M(1<=N,Mm=8) which indicate the size of the labyrinth. Then N lines follow, each line contains M integers. The array indicates the layout of the labyrinth. There are five integers which indicate the different type of area in the labyrinth: 0: The area is a wall, Ignatius should not walk on it. 1: The area contains nothing, Ignatius can walk on it. 2: Ignatius' start position, Ignatius starts his escape from this position. 3: The exit of the labyrinth, Ignatius' target position. 4: The area contains a Bomb-Reset-Equipment, Ignatius can delay the exploding time by walking to these areas.
Output
For each test case, if Ignatius can get out of the labyrinth, you should output the minimum time he needs, else you should just output -1.
Sample Input
3 3 3 2 1 1 1 1 0 1 1 3 4 8 2 1 1 0 1 1 1 0 1 0 4 1 1 0 4 1 1 0 0 0 0 0 0 1 1 1 1 4 1 1 1 3 5 8 1 2 1 1 1 1 1 4 1 0 0 0 1 0 0 1 1 4 1 0 1 1 0 1 1 0 0 0 0 3 0 1 1 1 4 1 1 1 1 1
Sample Output
4 -1 13
Author
Ignatius.L
bfs 的应用范围:
求最值问题.....属于盲目搜索的一种....
特点是所用的内存大相对于dfs....
代码如下:
1 #include<cstdio>
2 #include<iostream>
3 #include<deque>
4 #define maxn 10
5 #define SET 6
6 using namespace std;
7
8 int map[maxn][maxn],nn,mm;
9 int dir[4][2]=
10 {
11 {0,1}, /*向右*/
12 {0,-1}, /*向左*/
13 {-1,0}, /*向下*/
14 {1,0} /*向上*/
15 } ;
16 struct node{
17 int x,y; /*记录位置*/
18 int ans,time; /*步数和时间*/
19 }start;
20
21 /*生成地图*/
22 void save_map()
23 {
24 for(int i=0;i<nn;i++)
25 {
26 for(int j=0;j<mm;j++)
27 {
28 scanf("%d",&map[i][j]);
29 if(map[i][j]==2)
30 {
31 start.x=i;
32 start.y=j;
33 start.ans=0;
34 start.time=SET; /*设定为6s*/
35 }
36 }
37 }
38 return ;
39 }
40
41 void bfs( void )
42 {
43 deque<node>q ; /*队列实现*/
44
45 node q1,q2; //暂存
46 q.push_back(start); //将start的位置初始化.....
47
48 /* 当队列不为空的时候,执行下列程序 */
49 while(!q.empty())
50 {
51 q1=q.front(); //将对头的数据拿出来--->q1;
52 q.pop_front(); //q.pop()一样
53 for(int i=0; i<4 ;i++)
54 {
55 /*进入下一个状态*/
56 q2.x=q1.x+dir[i][0];
57 q2.y=q1.y+dir[i][1];
58 q2.ans=q1.ans+1;
59 q2.time=q1.time-1;
60 //如果满足这些条件便可进行下一步搜索
61 if(q2.x>=0&&q2.y>=0&&q2.x<nn&&q2.y<=mm&&map[q2.x][q2.y]!=0&&q2.time>0)
62 {
63 /*说明找到了答案,可以结束了*/
64 if(map[q2.x][q2.y]==3)
65 {
66 printf("%dn",q2.ans);
67
68 return ;
69 }
70 else if(map[q2.x][q2.y]==4)
71 {
72 /*重置时间,其他照常*/
73 q2.time=SET;
74 map[q2.x][q2.y]=0; //走过,不可再走,不然没完没了
75 }
76 q.push_back(q2);
77 }
78 }
79 }
80 /*说明不存在*/
81 printf("-1n");
82 return ;
83 }
84
85 int main()
86 {
87 int test;
88 scanf("%d",&test);
89 while(test--)
90 {
91 scanf("%d %d",&nn,&mm);
92 save_map();
93 bfs();
94 }
95 return 0;
96 }
- JavaScript 教程
- JavaScript 编辑工具
- JavaScript 与HTML
- JavaScript 与Java
- JavaScript 数据结构
- JavaScript 基本数据类型
- JavaScript 特殊数据类型
- JavaScript 运算符
- JavaScript typeof 运算符
- JavaScript 表达式
- JavaScript 类型转换
- JavaScript 基本语法
- JavaScript 注释
- Javascript 基本处理流程
- Javascript 选择结构
- Javascript if 语句
- Javascript if 语句的嵌套
- Javascript switch 语句
- Javascript 循环结构
- Javascript 循环结构实例
- Javascript 跳转语句
- Javascript 控制语句总结
- Javascript 函数介绍
- Javascript 函数的定义
- Javascript 函数调用
- Javascript 几种特殊的函数
- JavaScript 内置函数简介
- Javascript eval() 函数
- Javascript isFinite() 函数
- Javascript isNaN() 函数
- parseInt() 与 parseFloat()
- escape() 与 unescape()
- Javascript 字符串介绍
- Javascript length属性
- javascript 字符串函数
- Javascript 日期对象简介
- Javascript 日期对象用途
- Date 对象属性和方法
- Javascript 数组是什么
- Javascript 创建数组
- Javascript 数组赋值与取值
- Javascript 数组属性和方法
- Python函数
- Effective C++条款3 我可以不使用const?
- t想成为微信斗图之王么?你需要这款开源工具的力量!
- 蛋糕被切成了几块
- 偿还技术债(1)-EventBus源码详解
- 两万六千字带你Kotlin入门
- 从源码看 Jetpack(7)-SavedStateHandle源码详解
- 从源码看 Jetpack(6)-ViewModel源码详解
- linux配置c++11编译环境
- Java 多线程编程(聊聊线程池)
- Java 多线程编程(“锁”事碎碎念)
- Spring Cloud Alibaba技术栈(下)
- Electron安装过程深入解析(读完此文解决Electron安装失败导致的无法启动,无法打包的问题)
- Kafka中副本机制的设计和原理
- Cocoapods更新出错