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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!".
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5 5
....E
.....
.....
##...
S#...
5 8
........
........
..A....A
A######.
S......E
|
|
Please help BH!
12
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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 using namespace std; 5 6 #define maxn 101 7 #define inf 0x3f3f3f3f 8 9 int dir[][2] = { 1, 0, 0, 1, -1, 0, 0, -1 }; 10 int dp[maxn][maxn][6]; 11 int sx, sy, ex, ey, n, m; 12 char map[maxn][maxn]; 13 14 struct node{ 15 int x, y, step, speed;//spead加速 16 }; 17 void bfs(){ 18 node now, next; 19 now.x = sx, now.y = sy, now.step = 0, now.speed = 0; 20 dp[sx][sy][0] = 0; 21 queue<node>Q; 22 Q.push(now); 23 while (!Q.empty()){ 24 now = Q.front(); Q.pop(); 25 for (int i = 0; i < 4; i++){ 26 next = now; 27 next.x += dir[i][0]; 28 next.y += dir[i][1]; 29 if (next.x < 0 || next.x >= n || next.y < 0 || next.y >= m || map[next.x][next.y] == ‘#‘)continue;//不可行状态 30 if (next.speed){ 31 //加速效果 32 next.speed--; 33 next.step++; 34 } 35 else next.step += 2; 36 if (map[next.x][next.y] == ‘A‘)next.speed = 5;//获得加速神符 37 if (next.step < dp[next.x][next.y][next.speed]){ 38 dp[next.x][next.y][next.speed] = next.step; 39 Q.push(next); 40 } 41 } 42 } 43 int ans = inf; 44 for (int i = 4; i >= 0; i--) 45 ans = min(ans, dp[ex][ey][i]); 46 if (ans >= inf) 47 cout << "Please help BH!\r\n"; 48 else 49 cout << ans << "\r\n"; 50 } 51 int main(){ 52 while (cin >> n >> m){ 53 memset(dp, inf, sizeof dp); 54 for (int i = 0; i < n; i++){ 55 cin >> map[i]; 56 for (int j = 0; j < m; j++){ 57 if (map[i][j] == ‘S‘)sx = i, sy = j; 58 if (map[i][j] == ‘E‘)ex = i, ey = j; 59 } 60 } 61 bfs(); 62 } 63 return 0; 64 }
[Swust OJ 1023]--Escape(带点其他状态的BFS)
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原文地址：http://www.cnblogs.com/zyxStar/p/4593378.html
