解题心得：
1、point：关于可以返回路径的BFS的标记方法，并非是简单的0-1，而是可以用时间比较之后判断是否push。
2、queue创建的地点（初始化问题），在全局中创建queue在一次调用BFS()函数之后并不会初始化，应该在BFS()函数之中创建queue。
3、有关BFS的题可以用不同的函数实现不同的功能（maps_store、check、BFS），在查找bug的时候思路更为清晰。
4、边界一定要找清楚，怎么处理边界，尽量不重复，不超过，边界可以check中进行处理，这样check之后以免忘记。(尽量在开始的时候就把边界清楚的找出来，不然在后面会出现各种bug，并且很难找出)。
说明：（转） 这题不同于其他题的地方就是于虽然也是bfs，但对于走过的路径不能标记，因为可能还要走，注意题目要求：如果可以，可以走任意多遍 这就引发了一个问题，如果不缩减搜索范围，怎么可能走得出来呢？应该说这题好就好在不是根据走过的路径来标记，而是根据前后两次踏上同一位置是bomb离explode的时间长短来标记。简言之，如果第二次踏上一个位置，那么找出路已用的时间肯定是增加了，那为啥还要走上这条路呢？唯一的追求就是bomb离爆炸的时间增大了。所以可以利用这个条件来标记了。每次在入队前检查下爆炸时间是否比上次在同一位置的大，若是，则入队；反之，入队无意义了。详见代码一。从以上的分析中可以引出另一思路，也就是只要进入位置4，那么bomb就会延时到6分钟，最大的延时时间。换句话说，下次再进入该4位置，也不会获得更大的延时时间了。所以，只要访问过位置4了，就可以直接标记为0位置，表明下次不可在访问。
题目：
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
#include<stdio.h>
#include<queue>
#include<cstring>
using namespace std;
int t_remain,t_use;//记录步骤和剩余的时间，t_remain是多余的。
int row,col;//行列
int maps[10][10],use[10][10];
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};//方向控制
struct node//记录每一个点的位置和用的时间和剩余的时间，尽量记录完全
{
int x,y;
int t_remain;
int t_use;
};
node now,Next;//标配
bool flag;//用于标记找到目的地
void maps_store()//输入函数打包
{
memset(use,0,sizeof(use));
for(int i=0;i<row;i++)
for(int j=0;j<col;j++)
{
scanf("%d",&maps[i][j]);
if(maps[i][j]==2)
{
now.x = i;
now.y = j;
now.t_use = 0;
now.t_remain = 6;
use[now.x][now.y] = 6;
}
}
}
int check()//判断边界及不可以通过的点
{
if(Next.x<0 || Next.y <0 || Next.x>=row || Next.y>=col || maps[Next.x][Next.y]==0)
return 0;
else
return 1;
}
void BFS()//核心函数
{
queue <node> qu;//创建的地点一定要弄清楚（有关初始化）（找瞎了眼）。
qu.push(now);
while((!qu.empty()) && (!flag))
{
now = qu.front();
qu.pop();
for(int i=0;i<4;i++)
{
Next.x = now.x + dir[i][0];
Next.y = now.y + dir[i][1];
Next.t_use = now.t_use + 1;
Next.t_remain = now.t_remain - 1;
if(check())
{
if(maps[Next.x][Next.y] == 3)
{
t_use = Next.t_use;
t_remain = Next.t_remain;
flag = 1;
break;
}
if(maps[Next.x][Next.y] == 4)
{
Next.t_remain = 6;
}
if(Next.t_remain > use[Next.x][Next.y] && Next.t_remain>1)//核心技术：Next.t_remain > use[Next.x][Next.y]，这是用来判断是否回走。
{
use[Next.x][Next.y] = Next.t_remain;
qu.push(Next);
}
}
}
if(flag)
break;
}
return;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
flag = 0;
scanf("%d%d",&row,&col);
maps_store();
BFS();
if(flag && t_remain>0)
printf("%d
",t_use);
else
printf("%d
",-1);
}
return 0;
}