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

题目链接

在地图中从起点走向终点，起始时间为6，走一步消耗时间1，特殊点位可以重置时间，时间耗尽失败，求最短时间。

这道题目适合用bfs（宽度优先搜索），但是我用的是dfs（深度优先搜索）。

在搜索中更新到达搜索点位的步数和拥有时间，在之后的搜索中若搜索到更大的步数到达此点或者到达此点是拥有更少的时间则剪枝。

AC代码:

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
#define XDebug(x) cout<<#x<<"="<<x<<endl;
#define ArrayDebug(x,i) cout<<#x<<"["<<i<<"]="<<x[i]<<endl;
#define print(x) out(x);putchar('\n')
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int INF = 0x3f3f3f3f;
const int maxn = 10;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
template <class T>
inline bool read(T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) {
        return 0;
    }
    while (c != '-' && (c < '0' || c > '9')) {
        c = getchar();
    }
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0' && c <= '9') {
        ret = ret * 10 + (c - '0');
    }
    ret *= sgn;
    return 1;
}
template <class T>
inline void out(T x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) {
        out(x / 10);
    }
    putchar(x % 10 + '0');
}

int T;
int n, m;
int StartX, StartY;
int EndX, EndY;
int ans;
int plat[maxn][maxn];
int step[maxn][maxn];
int Time[maxn][maxn];

void dfs(int x, int y, int cnt, int time) {
	// 时间耗完返回；当前步数已经大于记录结果，剪枝
	if (time == 0 || cnt >= ans) {
		return;
	}
	// 记录结果
	if (x == EndX && y == EndY) {
		ans = min(ans, cnt);
		return;
	}
	// 重置
	if (plat[x][y] == 4) {
		time = 6;
	}
	// 若利用过更少的步数到达此点或者有过到达过此点时拥有的时间更长，剪枝
	if (step[x][y] <= cnt && time <= Time[x][y]) {
		return;
	}
	// 记录到达此点的步数和拥有时间
	step[x][y] = cnt;
	Time[x][y] = time;
	for (int i = -1; i <= 1; ++i) {
		for (int j = -1; j <= 1; ++j) {
			if (abs(i) == abs(j)) {
				continue;
			}
			int nx = x + i, ny = y + j;
			if (nx >= 0 && nx < n && ny >= 0 && ny < m && plat[nx][ny] != 0) {
				dfs(nx, ny, cnt + 1, time - 1);
			}
		}
	}
}

int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
   freopen("in.txt", "r", stdin);
   freopen("out.txt", "w", stdout);
#endif
	read(T);
	for (int Case = 1; Case <= T; ++Case) {
		read(n); read(m);
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < m; ++j) {
				step[i][j] = INF;
				Time[i][j] = 0;
				read(plat[i][j]);
				if (plat[i][j] == 2) {
					StartX = i;
					StartY = j;
				}
				else if (plat[i][j] == 3) {
					EndX = i;
					EndY = j;
				}
			}
		}
		ans = INF;
		step[StartX][StartY] = 0;
		dfs(StartX, StartY, 0, 6);
		ans = ans == INF ? -1 : ans;
		print(ans);
	}
#ifndef ONLINE_JUDGE
   fclose(stdin);
   fclose(stdout);
   system("gedit out.txt");
#endif
    return 0;
}