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
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.
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
-1
13
#include<cstdio>
#include<iostream>
#include<deque>
#define maxn 10
#define SET 6
using namespace std; int map[maxn][maxn],nn,mm;
int dir[][]=
{
{,}, /*向右*/
{,-}, /*向左*/
{-,}, /*向下*/
{,} /*向上*/
} ;
struct node{
int x,y; /*记录位置*/
int ans,time; /*步数和时间*/
}start; /*生成地图*/
void save_map()
{
for(int i=;i<nn;i++)
{
for(int j=;j<mm;j++)
{
scanf("%d",&map[i][j]);
if(map[i][j]==)
{
start.x=i;
start.y=j;
start.ans=;
start.time=SET; /*设定为6s*/
}
}
}
return ;
} void bfs( void )
{
deque<node>q ; /*队列实现*/ node q1,q2; //暂存
q.push_back(start); //将start的位置初始化..... /* 当队列不为空的时候,执行下列程序 */
while(!q.empty())
{
q1=q.front(); //将对头的数据拿出来--->q1;
q.pop_front(); //q.pop()一样
for(int i=; i< ;i++)
{
/*进入下一个状态*/
q2.x=q1.x+dir[i][];
q2.y=q1.y+dir[i][];
q2.ans=q1.ans+;
q2.time=q1.time-;
//如果满足这些条件便可进行下一步搜索
if(q2.x>=&&q2.y>=&&q2.x<nn&&q2.y<=mm&&map[q2.x][q2.y]!=&&q2.time>)
{
/*说明找到了答案,可以结束了*/
if(map[q2.x][q2.y]==)
{
printf("%d\n",q2.ans); return ;
}
else if(map[q2.x][q2.y]==)
{
/*重置时间,其他照常*/
q2.time=SET;
map[q2.x][q2.y]=; //走过,不可再走,不然没完没了
}
q.push_back(q2);
}
}
}
/*说明不存在*/
printf("-1\n");
return ;
} int main()
{
int test;
scanf("%d",&test);
while(test--)
{
scanf("%d %d",&nn,&mm);
save_map();
bfs();
}
return ;
}
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