[Swust OJ 1023]--Escape(带点其他状态的BFS)

 

解题思路：http://acm.swust.edu.cn/problem/1023/

Time limit(ms): 5000　　　　　　　　Memory limit(kb): 65535

 

 

Description

BH is in a maze,the maze is a matrix,he wants to escape!

 

Input

The input consists of multiple test cases.

For each case,the first line contains 2 integers N,M( 1 <= N, M <= 100 ).

Each of the following N lines contain M characters. Each character means a cell of the map.

Here is the definition for chracter.

For a character in the map: 

'S':BH's start place,only one in the map.

'E':the goal cell,only one in the map.

'.':empty cell.

'#':obstacle cell.

'A':accelerated rune.

BH can move to 4 directions(up,down,left,right) in each step.It cost 2 seconds without accelerated rune.When he get accelerated rune,moving one step only cost 1 second.The buff lasts 5 seconds,and the time doesn't stack when you get another accelerated rune.(that means in anytime BH gets an accelerated rune,the buff time become 5 seconds).

 

 

Output

The minimum time BH get to the goal cell,if he can't,print "Please help BH!".

 

Sample Input

5 5 ....E ..... ..... ##... S#...   5 8 ........ ........ ..A....A A######. S......E

Sample Output

Please help BH! 12

 

由于OJ上传数据的BUG,换行请使用"\r\n"，非常抱歉

 

题目大意：一个迷宫逃离问题，只是有了加速符A,正常情况下通过一个格子2s,有了加速符只要1s,并且加速符持续5s,‘S'代表起点

　　　　　'E'代表终点，'#'代表障碍，'.'空格子，能够逃离输出最少用时，否则输出"Please help BH!"

解题思路：BFS，用一个3维dp数组存贮，每一点在不同加速状态下的值，然后筛选dp数组终点的最小值即可

代码如下：

 1 #include <iostream>

 2 #include <cstring>

 3 #include <queue>

 4 #include <algorithm>

 5 using namespace std;

 6 

 7 #define maxn 101

 8 #define inf 0x3f3f3f3f

 9 

10 int dir[][2] = { 1, 0, 0, 1, -1, 0, 0, -1 };

11 int dp[maxn][maxn][6];

12 int sx, sy, ex, ey, n, m;

13 char map[maxn][maxn];

14 

15 struct node{

16     int x, y, step, speed;//spead加速

17 };

18 void bfs(){

19     node now, next;

20     now.x = sx, now.y = sy, now.step = 0, now.speed = 0;

21     dp[sx][sy][0] = 0;

22     queue<node>Q;

23     Q.push(now);

24     while (!Q.empty()){

25         now = Q.front(); Q.pop();

26         for (int i = 0; i < 4; i++){

27             next = now;

28             next.x += dir[i][0];

29             next.y += dir[i][1];

30             if (next.x < 0 || next.x >= n || next.y < 0 || next.y >= m || map[next.x][next.y] == '#')continue;//不可行状态

31             if (next.speed){

32                 //加速效果

33                 next.speed--;

34                 next.step++;

35             }

36             else next.step += 2;

37             if (map[next.x][next.y] == 'A')next.speed = 5;//获得加速神符

38             if (next.step < dp[next.x][next.y][next.speed]){

39                 dp[next.x][next.y][next.speed] = next.step;

40                 Q.push(next);

41             }

42         }

43     }

44     int ans = inf;

45     for (int i = 4; i >= 0; i--)

46         ans = min(ans, dp[ex][ey][i]);

47     if (ans >= inf)

48         cout << "Please help BH!\r\n";

49     else

50         cout << ans << "\r\n";

51 }

52 int main(){

53     while (cin >> n >> m){

54         memset(dp, inf, sizeof dp);

55         for (int i = 0; i < n; i++){

56             cin >> map[i];

57             for (int j = 0; j < m; j++){

58                 if (map[i][j] == 'S')sx = i, sy = j;

59                 if (map[i][j] == 'E')ex = i, ey = j;

60             }

61         }

62         bfs();

63     }

64     return 0;

65 }

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