Saving Tang Monk II HihoCoder - 1828（2018北京网络赛三维标记+bfs）

《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.

Input
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.

Output
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

Sample Input
2 2
S#
#T
2 5
SB###
##P#T
4 7
SP…
P#…
…#
B…##T
0 0
Sample Output
-1
8
11
题意：孙悟空救唐僧，#代表着毒气室，只有有氧气管的时候才能进入，出了毒气室要休息一分钟。B代表着氧气室，最多可以带5瓶氧气。P代表着可以加速一分钟。. 代表着空闲地方。问孙悟空从s到达t的最少时间是多少。
之前还没做到过这样的题目，之前的广搜就是判断这个点有没有走到过，走到就不能走了。但是这个题要是这样做的话就不对了，也非常麻烦。我们用vis[x][y][num]代表着走到（x,y）还剩num瓶氧气。如果到达一个点它的氧气瓶的数量之前出现过，那么这种状态之前也出现过。广搜暴力去找所有的状态，直到找到了唐僧。以为在出了毒气室的时候，需要休息一分钟，就代表着进毒气室需要2分钟，这样的话就需要用到优先队列来维护了。
代码如下：

#include<stdio.h>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
char mp[105][105];
int vis[105][105][1005];
struct node{int x,y,num,step;node(int a,int b,int c,int d){x=a;y=b;num=c;step=d;}bool operator <(const node &a)const {return a.step<step;}
};
int n,m;
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int bfs(int sx,int sy){priority_queue<node>q;node k=node(sx,sy,0,0);q.push(k);vis[sx][sy][0]=1;while(!q.empty()){node t=q.top();q.pop();if(mp[t.x][t.y]=='T'){return t.step;}for(int i=0;i<4;i++){int tx=t.x+dir[i][0];int ty=t.y+dir[i][1];if(tx<0||ty<0||tx>=n||ty>=m)continue;if(mp[tx][ty]=='#'&&t.num>0&&!vis[tx][ty][t.num-1]){vis[tx][ty][t.num-1]=1;q.push(node(tx,ty,t.num-1,t.step+2));}else if(mp[tx][ty]=='B'&&!vis[tx][ty][t.num+1]&&t.num+1<=5){vis[tx][ty][t.num+1]=1;q.push(node(tx,ty,t.num+1,t.step+1));}else if(mp[tx][ty]=='B'&&!vis[tx][ty][t.num]&&t.num+1>5){//最多只能携带五瓶vis[tx][ty][t.num]=1;q.push(node(tx,ty,t.num,t.step+1));}else if(mp[tx][ty]=='P'&&!vis[tx][ty][t.num]){vis[tx][ty][t.num]=1;q.push(node(tx,ty,t.num,t.step));}else if((mp[tx][ty]=='.'||mp[tx][ty]=='S'||mp[tx][ty]=='T')&&!vis[tx][ty][t.num]){vis[tx][ty][t.num]=1;q.push(node(tx,ty,t.num,t.step+1));}}}return -1;
}
int main(){while(scanf("%d%d",&n,&m)!=EOF){if(n==0&&m==0)return 0;for(int i=0;i<n;i++){scanf("%s",mp[i]);}memset(vis,0,sizeof(vis));int sx,sy;for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(mp[i][j]=='S'){sx=i;sy=j;break;}}}printf("%d\n",bfs(sx,sy));}
}

努力加油a啊，(^o)/~

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