Nightmare
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4360 Accepted Submission(s): 2177
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.
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.
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
4 -1 13
//1072
#include <iostream>
#include <fstream>
#include <queue>
using namespace std;
typedef struct node {
int i;
int j;
int time;//到达这个点的时候的时间
int lifetime;//此时的生命时间值
friend bool operator < (node n1, node n2)
{
return n1.time > n2.time;
}
}Node;
typedef struct
{
int i;
int j;
}Point;
int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
int valbles[8][8];//表示上次插入的时候的剩余时间
char map[8][8];
int n, m;
Point start;//开始点
Point dest;//出口
int bfs()
{
if (start.i == dest.i && start.j == dest.j) return 0;
priority_queue<Node> Q;
Node temp1;
temp1.i = start.i;
temp1.j = start.j;
temp1.time = 0;
temp1.lifetime = 6;
Q.push(temp1);
//used[start.i][start.j] = true;//已经走过
valbles[start.i][start.j] = 6;
while(!Q.empty())
{
Node temp = Q.top();
Q.pop();
if (temp.lifetime == 1) continue;//已经没有生命数值
if (temp.time >= n*m)
return -1;
for (int i = 0; i < 4; ++ i)//扩展四个方向的路径
{
int ni = temp.i + dir[i][0];
int nj = temp.j + dir[i][1];
if (ni >= 0 && ni < n && nj >= 0 && nj < m && map[ni][nj] != '0')
{
if (map[ni][nj] == '3')//已经达到出口
return temp.time + 1;
Node extNode;
extNode.i = ni;
extNode.j = nj;
extNode.time = temp.time + 1;
if (map[ni][nj] == '1')//直接走，生命数值减去一
extNode.lifetime = temp.lifetime - 1;
else
if (map[ni][nj] == '4')//回复生命数值到6
extNode.lifetime = 6;
//used[ni][nj] = true; //已经走过
if (extNode.lifetime <= valbles[ni][nj])continue;//没有继续插入的价值
valbles[ni][nj] = extNode.lifetime;
Q.push(extNode);
}
}
}
return -1;
}
int main()
{
//ifstream cin("in.txt");
int num ;
cin >> num ;
for (int i =0; i < num; ++ i)
{
cin >> n >> m;
for (int j = 0; j < n; ++ j)
for (int k = 0; k < m ; ++ k)
{
cin >> map[j][k];
valbles[j][k] = -1;
if (map[j][k] == '2')
{
start.i = j;
start.j = k;
}
else
if (map[j][k] == '3')
{
dest.i = j;
dest.j = k;
}
}
int result = bfs();
cout << result << endl;
}
//cin.close();
return 0;
}