[Swust OJ 1023]--Escape(带点其他状态的BFS)
解题思路:http://acm.swust.edu.cn/problem/1023/
BH is in a maze,the maze is a matrix,he wants to escape!
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).
The minimum time BH get to the goal cell,if he can't,print "Please help BH!".
5 5
....E
.....
.....
##...
S#...
5 8
........
........
..A....A
A######.
S......E
|
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:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }