HDU 5025 Saving Tang Monk 【狀態壓縮BFS】
阿新 • • 發佈：2018-09-24
hat urn clear there this imp sha 初始化 沒有
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. But to rescue Tang Monk, Sun Wukong might need to get some keys and kill some snakes in his way.
The palace can be described as a matrix of characters. Each character stands for a room. In the matrix, ‘K‘ represents the original position of Sun Wukong, ‘T‘ represents the location of Tang Monk and ‘S‘ stands for a room with a snake in it. Please note that there are only one ‘K‘ and one ‘T‘, and at most five snakes in the palace. And, ‘.‘ means a clear room as well ‘#‘ means a deadly room which Sun Wukong couldn‘t get in.
There may be some keys of different kinds scattered in the rooms, but there is at most one key in one room. There are at most 9 kinds of keys. A room with a key in it is represented by a digit(from ‘1‘ to ‘9‘). For example, ‘1‘ means a room with a first kind key, ‘2‘ means a room with a second kind key, ‘3‘ means a room with a third kind key... etc. To save Tang Monk, Sun Wukong must get ALL kinds of keys(in other words, at least one key for each kind).
For each step, Sun Wukong could move to the adjacent rooms(except deadly rooms) in 4 directions(north, west, south and east), and each step took him one minute. If he entered a room in which a living snake stayed, he must kill the snake. Killing a snake also took one minute. If Sun Wukong entered a room where there is a key of kind N, Sun would get that key if and only if he had already got keys of kind 1,kind 2 ... and kind N-1. In other words, Sun Wukong must get a key of kind N before he could get a key of kind N+1 (N>=1). If Sun Wukong got all keys he needed and entered the room in which Tang Monk was cuffed, the rescue mission is completed. If Sun Wukong didn‘t get enough keys, he still could pass through Tang Monk‘s room. Since Sun Wukong was a impatient monkey, he wanted to save Tang Monk as quickly as possible. Please figure out the minimum time Sun Wukong needed to rescue Tang Monk.
For each case, the first line includes two integers N and M(0 < N <= 100, 0<=M<=9), meaning that the palace is a N×N matrix and Sun Wukong needed M kinds of keys(kind 1, kind 2, ... kind M).
Then the N × N matrix follows.
The input ends with N = 0 and M = 0. 
任意門：http://acm.hdu.edu.cn/showproblem.php?pid=5025
Saving Tang Monk
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3242 Accepted Submission(s): 1127
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. But to rescue Tang Monk, Sun Wukong might need to get some keys and kill some snakes in his way.
The palace can be described as a matrix of characters. Each character stands for a room. In the matrix, ‘K‘ represents the original position of Sun Wukong, ‘T‘ represents the location of Tang Monk and ‘S‘ stands for a room with a snake in it. Please note that there are only one ‘K‘ and one ‘T‘, and at most five snakes in the palace. And, ‘.‘ means a clear room as well ‘#‘ means a deadly room which Sun Wukong couldn‘t get in.
There may be some keys of different kinds scattered in the rooms, but there is at most one key in one room. There are at most 9 kinds of keys. A room with a key in it is represented by a digit(from ‘1‘ to ‘9‘). For example, ‘1‘ means a room with a first kind key, ‘2‘ means a room with a second kind key, ‘3‘ means a room with a third kind key... etc. To save Tang Monk, Sun Wukong must get ALL kinds of keys(in other words, at least one key for each kind).
For each step, Sun Wukong could move to the adjacent rooms(except deadly rooms) in 4 directions(north, west, south and east), and each step took him one minute. If he entered a room in which a living snake stayed, he must kill the snake. Killing a snake also took one minute. If Sun Wukong entered a room where there is a key of kind N, Sun would get that key if and only if he had already got keys of kind 1,kind 2 ... and kind N-1. In other words, Sun Wukong must get a key of kind N before he could get a key of kind N+1 (N>=1). If Sun Wukong got all keys he needed and entered the room in which Tang Monk was cuffed, the rescue mission is completed. If Sun Wukong didn‘t get enough keys, he still could pass through Tang Monk‘s room. Since Sun Wukong was a impatient monkey, he wanted to save Tang Monk as quickly as possible. Please figure out the minimum time Sun Wukong needed to rescue Tang Monk.
For each case, the first line includes two integers N and M(0 < N <= 100, 0<=M<=9), meaning that the palace is a N×N matrix and Sun Wukong needed M kinds of keys(kind 1, kind 2, ... kind M).
Then the N × N matrix follows.
The input ends with N = 0 and M = 0.
Sample Input 3 1 K.S ##1 1#T 3 1 K#T .S# 1#. 3 2 K#T .S. 21. 0 0
Sample Output 5 impossible 8
題意概括：
給一個 N * N 的地圖，孫悟空起點在 K ，唐僧起點在 T，數字 i 代表第 i 把鑰匙，# 是毒氣區不能進入， S 有蛇（需要殺死才能通過，只需要殺死一次）；
孫悟空要按順序集齊 M 把鑰匙才能解救師父，求最短的時間，如果沒有輸出“impossible”。
解題思路：
BFS找最短路，因為蛇最多只有5條，所以狀態壓縮判斷哪條已殺，哪條未殺；如果未殺則當前需要多走一步，反則不用。
集鑰匙只需要一個變量 cnt_key 記錄已經集了多少把鑰匙，那麽接下倆可以拿的鑰匙就是 第 cnt_key+1 把。
BFS需要一個 vis[ x ][ y ][ key ][ snake ] 來book一下狀態，已經處理過的不再處理。
最後一個debug了很久的原因：多測試用例當 N M 都為 0 時（ ... &&（N+M））才結束，習慣性打了（ ... &&N && M），涼涼。
AC code:
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cstring> 5 #include <queue> 6 #include <cmath> 7 #define INF 0x3f3f3f3f 8 using namespace std; 9 const int MAXN = 105; 10 int nx[] = {-1, 1, 0, 0}; 11 int ny[] = {0, 0, -1, 1}; 12 struct date 13 { 14 int x, y, t, key, snake; 15 date(int _x = 0, int _y = 0, int _t = 0, int _key = 0, int _snake = 0):x(_x), y(_y), t(_t), key(_key), snake(_snake){} 16 }; 17 18 char mmp[MAXN][MAXN]; 19 bool vis[MAXN][MAXN][10][40]; 20 int N, M, cnt; 21 int sx, sy, ex, ey; 22 23 void solve() 24 { 25 memset(vis, false, sizeof(vis)); 26 int ans = INF; //初始化步數 27 queue<date> sq; 28 sq.push(date(sx, sy, 0, 0, 0)); 29 //vis[sx][sy] = true; 30 date tp; 31 while(!sq.empty()){ 32 tp = sq.front(), sq.pop(); 33 int x = tp.x, y = tp.y, key = tp.key, snake = tp.snake, t = tp.t; 34 if(key == M && mmp[x][y] == ‘T‘) ans = min(ans, t); //集齊鑰匙並且到達終點 35 if(vis[x][y][key][snake]) continue; //狀態已訪問過 36 vis[x][y][key][snake] = true; //狀態不重復 37 for(int k = 0; k < 4; k++){ 38 int tx = x + nx[k], ty = y + ny[k]; 39 if(tx < 0 || ty < 0 || tx == N || ty == N || mmp[tx][ty] == ‘#‘) continue; 40 41 date now = tp; 42 43 if(‘A‘ <= mmp[tx][ty] && mmp[tx][ty] <= ‘G‘){ 44 int s = mmp[tx][ty] - ‘A‘; 45 if((1<<s) & now.snake); //蛇被被打了 46 else{ //蛇沒有被打 47 now.snake |= (1<<s); 48 now.t++; 49 } 50 } 51 else if(mmp[tx][ty] - ‘0‘ == now.key + 1){ //撿鑰匙，遇到當前可以撿的鑰匙 52 now.key++; 53 } 54 now.t++; 55 sq.push(date(tx, ty, now.t, now.key, now.snake)); 56 } 57 } 58 if(ans >= INF) puts("impossible"); 59 else printf("%d\n", ans); 60 } 61 62 int main() 63 { 64 while(~scanf("%d%d", &N, &M) && (N+M)){ 65 cnt = 0; 66 for(int i = 0; i < N; i++){ 67 scanf("%s", &mmp[i]); 68 for(int j = 0; j < N; j++){ 69 if(mmp[i][j] == ‘K‘) sx = i, sy = j; 70 if(mmp[i][j] == ‘S‘) mmp[i][j] = cnt+‘A‘, cnt++; 71 } 72 } 73 solve(); 74 } 75 return 0; 76 }View Code
HDU 5025 Saving Tang Monk 【狀態壓縮BFS】