UVA 1601 The Morning after Halloween(搜索,二维数组转为图)

  1 //#include<bits/stdc++.h>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<iostream>
  5 using namespace std;
  6 typedef long long ll;
  7 int n,m,k;
  8 char maps[20][20];
  9 int vis[200][200][200];
 10 int num[20][20];
 11 int que[10000000][4];
 12 int connect[200][200];
 13 int goal[4];
 14 int all;
 15 int ans = -1;
 16 void bfs(){
 17     int front = 0 , rear = 1;
 18     while(front < rear){
 19         int a = que[front][1] , b = que[front][2] , c = que[front][3];
 20         vis[a][b][c] = true;
 21         int step = que[front][0];
 22 
 23         //cout << a << " - " << goal[1] << endl;
 24         //cout << b << " - " << goal[2] << endl;
 25         //cout << c << " - " << goal[3] << endl;
 26 
 27         if(a == goal[1] && b == goal[2] && c == goal[3]){
 28             ans = step;
 29             return;
 30         }
 31         // a 是图上某个点  , coonect是邻接表。
 32         for(int i = 0 ; i <= connect[a][0] ; i ++){
 33             int na = i ? connect[a][i] : a;
 34             for(int j = 0 ; j <= connect[b][0] ; j ++){
 35                 int nb = j ? connect[b][j] : b;
 36                 for(int k = 0 ; k <= connect[c][0] ; k ++){
 37                     int nc = k ? connect[c][k] : c;
 38                     if((na && nb && na == nb) ||(na && nc && na == nc)||(nb && nc && nb == nc)) continue;
 39                     if((na && nb) && na == b && nb == a) continue;
 40                     if((nb && nc) && nb == c && nc == b) continue;
 41                     if((na && nc) && na == c && nc == a) continue;
 42                     if(!vis[na][nb][nc]){
 43                         vis[na][nb][nc] = true;
 44                         que[rear][0] = step + 1;
 45                         que[rear][1] = na;
 46                         que[rear][2] = nb;
 47                         que[rear][3] = nc;
 48                         rear ++;
 49                     }
 50                 }
 51             }
 52         }
 53         front ++;
 54     }
 55 }
 56 int main(){
 57     /**
 58      *先计算有几个点,  1......n
 59      *再遍历每个点 pos(1~n), 与之相邻的 建立邻接表
 60      * 再将起点 终点映射为 图中的下标。
 61      * */
 62     while(scanf("%d%d%d",&m,&n,&k) != EOF && (n||m||k)){
 63         getchar();
 64         for(int i = 0 ; i < n ; i ++){
 65             gets(maps[i]);
 66         }
 67         all = 0;
 68         memset(connect,0,sizeof(connect));
 69         memset(num,0,sizeof(num));
 70         memset(vis,0,sizeof(vis));
 71         for(int i = 0 ; i < n ; i ++){
 72             for(int j = 0 ; j < m ; j ++){
 73                 // 图的下标从一开始
 74                 if(maps[i][j] != ‘#‘) num[i][j] = ++all;
 75             }
 76         }
 77         for(int i = 0 ; i < n ; i ++){
 78             for(int j = 0 ; j < m ; j ++){
 79                 int &pos = num[i][j];
 80                 if(num[i][j]){
 81                     // 转为邻接表， connect[pos][0] 表示有几个点和该点相连。
 82                     if(num[i-1][j]) connect[pos][++connect[pos][0]] = num[i-1][j];
 83                     if(num[i+1][j]) connect[pos][++connect[pos][0]] = num[i+1][j];
 84                     if(num[i][j-1]) connect[pos][++connect[pos][0]] = num[i][j-1];
 85                     if(num[i][j+1]) connect[pos][++connect[pos][0]] = num[i][j+1];
 86                 }
 87             }
 88         }
 89 
 90         que[0][1] = que[0][2] = que[0][3] = goal[1] = goal[2] = goal[3] = 0;
 91         que[0][0] = 0;
 92         for(int i = 0 ; i < n ; i ++){
 93             for(int j = 0 ; j < m ; j ++){
 94                 if(num[i][j]){
 95                     if(maps[i][j] == ‘a‘) que[0][1] = num[i][j];
 96                     else if(maps[i][j] == ‘b‘) que[0][2] = num[i][j];
 97                     else if(maps[i][j] == ‘c‘) que[0][3] = num[i][j];
 98                     else if(maps[i][j] == ‘A‘) goal[1] = num[i][j];
 99                     else if(maps[i][j] == ‘B‘) goal[2] = num[i][j];
100                     else if(maps[i][j] == ‘C‘) goal[3] = num[i][j];
101                 }
102             }
103         }
104         bfs();
105         printf("%d\n",ans);
106     }
107     return 0;
108 }
109 /**
110 4 3 1
111 ####
112 #aA#
113 ####
114 5 5 2
115 #####
116 #A#B#
117 #   #
118 #b#a#
119 #####
120 16 4 3
121 ################
122 ## ########## ##
123 #    ABCcba    #
124 ################
125 16 16 3
126 ################
127 ### ##    #   ##
128 ##  #  ##   # c#
129 #  ## ########b#
130 # ##  # #   #  #
131 #  # ##   # # ##
132 ##  a#  # # #  #
133 ### ## #### ## #
134 ##   #   #  #  #
135 #  ##### # ## ##
136 ####   #B# #   #
137 ##  C#   #   ###
138 #  # # ####### #
139 # ######  A##  #
140 #        #    ##
141 ################
142 0 0 0
143 36
144 */