[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：

 #include <iostream>
 #include <cstring>
 #include <queue>
 #include <algorithm>
 using namespace std;

 #define maxn 101
 #define inf 0x3f3f3f3f

 int dir[][] = { , , , , -, , , - };
 int dp[maxn][maxn][];
 int sx, sy, ex, ey, n, m;
 char map[maxn][maxn];

 struct node{
     int x, y, step, speed;//spead加速
 };
 void bfs(){
     node now, next;
     now.x = sx, now.y = sy, now.step = , now.speed = ;
     dp[sx][sy][] = ;
     queue<node>Q;
     Q.push(now);
     while (!Q.empty()){
         now = Q.front(); Q.pop();
         for (int i = ; i < ; i++){
             next = now;
             next.x += dir[i][];
             next.y += dir[i][];
             if (next.x <  || next.x >= n || next.y <  || next.y >= m || map[next.x][next.y] == '#')continue;//不可行状态
             if (next.speed){
                 //加速效果
                 next.speed--;
                 next.step++;
             }
             else next.step += ;
             if (map[next.x][next.y] == 'A')next.speed = ;//获得加速神符
             if (next.step < dp[next.x][next.y][next.speed]){
                 dp[next.x][next.y][next.speed] = next.step;
                 Q.push(next);
             }
         }
     }
     int ans = inf;
     for (int i = ; i >= ; i--)
         ans = min(ans, dp[ex][ey][i]);
     if (ans >= inf)
         cout << "Please help BH!\r\n";
     else
         cout << ans << "\r\n";
 }
 int main(){
     while (cin >> n >> m){
         memset(dp, inf, sizeof dp);
         for (int i = ; i < n; i++){
             cin >> map[i];
             for (int j = ; j < m; j++){
                 if (map[i][j] == 'S')sx = i, sy = j;
                 if (map[i][j] == 'E')ex = i, ey = j;
             }
         }
         bfs();
     }
     return ;
 }