題目
《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
題目大意
給你一個迷宮，S是起點，T是終點，要從S走到T，判斷最少的時間，無法走出迷宮則輸出-1。
中間#表示毒氣室，進入毒氣室必須有氧氣瓶，當進入毒氣室需要逗留一個時間單位。B表示的地方則爲有氧氣瓶的房間，每次經過可以拿一個氧氣瓶，手中的氧氣瓶最多不超過5個，‘.’就代表普通的路，P代表加速藥丸，可以加速一個時間單位。
我的理解
我的想法就是跑一個bfs，因爲每個或節點擴展的下一個節點的時間未必是增加的。所以用一個優先隊列作爲bfs的隊列。
而判斷一個點是否走過的vis數組不只是簡單的兩位數組記錄走過的點的位置，而是增加一個第三維度，代表氧氣瓶的個數，因爲當走到同一個點時，但手中的氧氣瓶個數不相同的時候，是不同的狀態，需要不同的記錄。
然後就是一個正常的bfs了。
代碼
#include <bits/stdc++.h>
using namespace std;
const int maxn = 110;
int vis[maxn][maxn][6];
char mp[maxn][maxn];
int n,m;
int dir[4][2] = {0,1,1,0,0,-1,-1,0};
struct node{
int x,y;
int ox;
int dis;
bool operator<(node obj)const{
return dis>obj.dis;
}
node(){}
node(int X,int Y,int Ox,int Dis){
x = X;
y = Y;
ox = Ox;
dis = Dis;
}
};
int bfs(int sx,int sy){
priority_queue<node> pq;
pq.push(node(sx,sy,0,0));
vis[sx][sy][0] = 1;
while(pq.size()){
node now = pq.top();
pq.pop();
// cout<<endl<<"& "<<now.x<<" "<<now.y<<" "<<now.ox<<" "<<now.dis<<endl;
if(mp[now.x][now.y] == 'T'){
return now.dis;
}
for(int i = 0;i<4;i++){
int tx = now.x+dir[i][0];
int ty = now.y+dir[i][1];
if(tx<1 || ty<1 || tx>n || ty>m){
continue;
}
int tox = now.ox;
// cout<<tx<<" "<<ty<<" "<<tox<<" "<<now.dis<<endl;
if(vis[tx][ty][tox]) continue;
vis[tx][ty][tox] = 1;
if(mp[tx][ty]=='B'){
pq.push(node(tx,ty,min(5,tox+1),now.dis+1));
}else if(mp[tx][ty] == '#'){
if(tox<=0) continue;
pq.push(node(tx,ty,tox-1,now.dis+2));
}else if(mp[tx][ty] == 'P'){
pq.push(node(tx,ty,tox,now.dis));
}else{
pq.push(node(tx,ty,tox,now.dis+1));
}
}
}
return -1;
}
int main(){
int sx,sy;
while(scanf("%d%d",&n,&m) && n+m){
memset(vis,0,sizeof(vis));
for(int i = 1;i<=n;i++){
scanf("%s",mp[i]+1);
}
bool fdst = false;
for(int i = 1;i<=n;i++){
for(int j = 1;j<=m;j++){
if(mp[i][j] == 'S'){
sx = i;
sy = j;
fdst = true;
break;
}
}
if(fdst) break;
}
int res = bfs(sx,sy);
printf("%d\n",res);
}
return 0;
}