題目鏈接：
http://acm.nyist.net/JudgeOnline/problem.php?pid=483
描述
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.
輸入
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.
輸出
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.
樣例輸入
| 2 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 |
樣例輸出
| 4 -1 |
題目中文描述
簡單翻一下就是給一個二維數組，從2開始，3結束，規定最多能走5步，每次只能從上下左右移動一步，1表示有路，0表示牆（不能走），4表示“補血”走的步數置0。問最少需要多少步可以走到出口，如若不能，輸出1.
算法思想：
因爲要遍歷所有的路徑，這樣才能尋找到最少的步數，故使用深度搜索。唯一需要特殊考慮的是遇到4的時候，需要將血補滿。這裏還需要使用一個二維數組將每個點所走的最少步數記錄了下來，深度搜索，填充該表即可。
注意：網上有些可以AC的代碼使用的是廣度搜索，是帶有一個能回溯的廣度搜索，有些是存在問題的，個人測試了一下，當遇到有多個4的時候，其中有些4是必須要走，有些4是不需要走的，這個時候廣度優先是會有問題的，之所以能AC，說明測試用例太少。
源代碼
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int map[10][10], Time[10][10], N, M, ans;
int dir[4][2] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
void DFS(int x, int y, int n, int time)
{
if (Time[x][y] >= time)
return;
if (!map[x][y])
return;
if (map[x][y] == 3)
{
ans = min(ans,n);
return;
}
if (map[x][y] == 4)
time = 6;
int temp = Time[x][y];
Time[x][y] = time;
n++;
int r, c;
for (int i = 0; i < 4; i++)
{
r = x + dir[i][0], c = y + dir[i][1];
DFS(r, c, n, time - 1);
}
Time[x][y] = temp;
}
int main()
{
int T;
int x, y;
cin >> T;
while (T--)
{
cin >> N >> M;
ans = 9999999;
memset(map, 0, sizeof(map));
memset(Time, 0, sizeof(Time));
for (int i = 1; i <= N; i++)
{
for (int j = 1; j <= M; j++)
{
cin >> map[i][j];
if (map[i][j] == 2)
x = i, y = j;
}
}
DFS(x, y, 0, 6);
if (ans == 9999999)
cout << "-1" << endl;
else
cout << ans << endl;
}
return 0;
}