描述

《Journey to the West》(also 《Monkey》) is one of the Four Great Classical Novels of Chinese literature. It was written by Wu Cheng‘en during the Ming Dynasty. In this novel, Monkey King Sun Wukong, pig Zhu Bajie and Sha Wujing, escorted Tang Monk to India to get sacred Buddhism texts.

During the journey, Tang Monk was often captured by demons. Most of demons wanted to eat Tang Monk to achieve immortality, but some female demons just wanted to marry him because he was handsome. So, fighting demons and saving Monk Tang is the major job for Sun Wukong to do.

Once, Tang Monk was captured by the demon White Bones. White Bones lived in a palace and she cuffed Tang Monk in a room. Sun Wukong managed to get into the palace, and he wanted to reach Tang Monk and rescue him.

The palace can be described as a matrix of characters. Different characters stand for different rooms as below:

‘S‘ : The original position of Sun Wukong

‘T‘ : The location of Tang Monk

‘.‘ : An empty room

‘#‘ : A deadly gas room.

‘B‘ : A room with unlimited number of oxygen bottles. Every time Sun Wukong entered a ‘B‘ room from other rooms, he would get an oxygen bottle. But staying there would not get Sun Wukong more oxygen bottles. Sun Wukong could carry at most 5 oxygen bottles at the same time.

‘P‘ : A room with unlimited number of speed-up pills. Every time Sun Wukong entered a ‘P‘ room from other rooms, he would get a speed-up pill. But staying there would not get Sun Wukong more speed-up pills. Sun Wukong could bring unlimited number of speed-up pills with him.

Sun Wukong could move in the palace. For each move, Sun Wukong might go to the adjacent rooms in 4 directions(north, west,south and east). But Sun Wukong couldn‘t get into a ‘#‘ room(deadly gas room) without an oxygen bottle. Entering a ‘#‘ room each time would cost Sun Wukong one oxygen bottle.

Each move took Sun Wukong one minute. But if Sun Wukong ate a speed-up pill, he could make next move without spending any time. In other words, each speed-up pill could save Sun Wukong one minute. And if Sun Wukong went into a ‘#‘ room, he had to stay there for one extra minute to recover his health.

Since Sun Wukong was an impatient monkey, he wanted to save Tang Monk as soon as possible. Please figure out the minimum time Sun Wukong needed to reach Tang Monk.

輸入

There are no more than 25 test cases.

For each case, the first line includes two integers N and M(0 < N,M ≤ 100), meaning that the palace is a N × M matrix.

Then the N×M matrix follows.

The input ends with N = 0 and M = 0.

輸出

For each test case, print the minimum time (in minute) Sun Wukong needed to save Tang Monk. If it‘s impossible for Sun Wukong to complete the mission, print -1

樣例輸入

2 2
S#
#T
2 5
SB###
##P#T
4 7
SP.....
P#.....
......#
B...##T
0 0

樣例輸出

-1
8
11

題意 ： 題目大意 給一個圖S是起點 T是終點 .是空房間 #是毒氣室 B是氧氣瓶存放室 P是加速室 每次走到空房間或者起點消耗1秒 走到氧氣室獲得一個氧氣瓶最多攜帶5個氧氣瓶 進入毒氣室需要一瓶氧氣並且消耗2秒 進入加速室不消耗時間(可以這麽理解) 問S走到T的最短時間

思路分析 ： 正常的分層廣搜即可，有個地方坑死我了，就是從S的位置去搜索，但是在搜的時候判斷條件裏忘記加 S 的判斷條件了，只要在開始廣搜的時候講 S 換成 . 即可

代碼示例：

using namespace std;
#define ll long long
int n, m;
char mp[105][105];
int sx, sy;

struct node
{
    int x, y;
    int yang, bu;
    bool operator< (const node &v)const {
        return bu > v.bu;
    }    
};
priority_queue<node>que;
bool vis[105][105][10];
int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};

bool check(int x, int y){
    if (x >= 1 && x <= n && y >= 1 && y <= m) return true;
    return false;
}

void bfs(){
    while(!que.empty()) que.pop();
    memset(vis, false, sizeof(vis));
    
    que.push({sx, sy, 0, 0});    
    node f;
    mp[sx][sy] = ‘.‘; 
    while(!que.empty()){
        node v = que.top();
        que.pop();
        
        if (mp[v.x][v.y] == ‘T‘) {
            printf("%d\n", v.bu);
            return;
        }
        if (vis[v.x][v.y][v.yang]) continue;
        vis[v.x][v.y][v.yang] = 1;
        
        for(int i = 0; i < 4; i++){
            int fx = v.x+dir[i][0];
            int fy = v.y+dir[i][1];
            if (check(fx, fy)){
                if (mp[fx][fy] == ‘.‘ || mp[fx][fy] == ‘T‘) {
                    if (!vis[fx][fy][v.yang]) que.push({fx, fy, v.yang, v.bu+1});
                }
                else if (mp[fx][fy] == ‘#‘){
                    if (v.yang == 0) continue;
                    if (!vis[fx][fy][v.yang-1]) que.push({fx, fy, v.yang-1, v.bu+2});
                }
                else if (mp[fx][fy] == ‘P‘){
                    if (!vis[fx][fy][v.yang]) que.push({fx, fy, v.yang, v.bu});
                }
                else if (mp[fx][fy] == ‘B‘){ 
                    if (v.yang+1 <= 5 && !vis[fx][fy][v.yang+1]) que.push({fx, fy, v.yang+1, v.bu+1});
                }
            } 
        }       
    }
    printf("-1\n");
}

int main() {

    while(~scanf("%d%d", &n, &m) && m+n){
        
        for(int i = 1; i <= n; i++){
            scanf("%s", mp[i]+1);
            for(int j = 1; j <= m; j++){
                if (mp[i][j] == ‘S‘) sx = i, sy = j;
            }
        }
        
        bfs();
    }
    return 0;
}

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