J
19
Jailbreak
John is on a mission to get two people out of prison. This particular prison is a one-story
building. He has managed to get hold of a detailed floor plan, indicating all the walls and
doors. He also knows the locations of the two people he needs to set free. The prison guards
are not the problem – he has planned a diversion that should leave the building practically
void.
The doors are his main concern. All doors are normally opened remotely from a control
room, but John can open them by other means. Once he has managed to open a door, it
remains open. However, opening a door takes time, which he does not have much of, since
his diversion will only work for so long. He therefore wants to minimize the number of doors
he needs to open. Can you help him plan the optimal route to get to the two prisoners?
Input
On the first line one positive number: the number of test cases, at most 100. After that per test
case:
• one line with two space-separated integers h and w (2 ≤ h, w ≤ 100): the width and
height of the map.
• h lines with w characters describing the prison building:
–
–
–
–
‘.’ is an empty space.
‘ * ’ is an impenetrable wall.
‘#’ is a door.
‘$’ is one of the two people to be liberated.
John can freely move around the outside of the building. There are exactly two people on the
map. For each person, a path from the outside to that person is guaranteed to exist.
Output
Per test case:
• one line with a single integer: the minimum number of doors John needs to open in
order to get to both prisoners.20
Problem J: Jailbreak
Sample in- and output
Input Output
3
5 9
****#****
*..#.#..*
****.****
*$#.#.#$*
*********
5 11
*#*********
*$*...*...*
*$*.*.*.*.*
*...*...*.*
*********.*
9 9
*#**#**#*
*#**#**#*
*#**#**#*
*#**.**#*
*#*#.#*#*
*$##*##$*
*#*****#*
*.#.#.#.*
********* 4
0
9
我的思路是这样的
先从两个人质出发做两遍bfs,做bfs的时候要记录路径
由于出口最多有400个(不到400)
然后枚举两个人质都能到达的出口
从两个人质到达该出口有两条路径
然后标记这两条路径上的门(注意重复的只标记一次)
然后统计门的数量.
然而wa了....
并不明白为什么.
改日再说.
1 /*************************************************************************
2 > File Name: code/whust/#0.2/J.cpp
3 > Author: 111qqz
4 > Email: [email protected]
5 > Created Time: 2015年08月09日 星期日 12时54分56秒
6 ************************************************************************/
7 #include<iostream>
8 #include<iomanip>
9 #include<cstdio>
10 #include<algorithm>
11 #include<cmath>
12 #include<cstring>
13 #include<string>
14 #include<map>
15 #include<set>
16 #include<queue>
17 #include<vector>
18 #include<stack>
19 #define y0 abc111qqz
20 #define y1 hust111qqz
21 #define yn hez111qqz
22 #define j1 cute111qqz
23 #define tm crazy111qqz
24 #define lr dying111qqz
25 using namespace std;
26 #define REP(i, n) for (int i=0;i<int(n);++i)
27 typedef long long LL;
28 typedef unsigned long long ULL;
29 const int inf = 0x7fffffff;
30 const int N=1E2+5;
31 char maze[N][N];
32 int tar[5][2];
33 int beg[4*N][2];
34 int dx[4]={0,0,-1,1};
35 int dy[4]={-1,1,0,0};
36 int total;
37 int h,w;
38 int sum;
39 bool v[N][N];
40 struct NODE
41 {
42 int d,prex,prey;
43 }q[N][N][2];
44 bool ok ( int x,int y,int num)
45 {
46 if (x>=0&&x<h&&y>=0&&y<w&&q[x][y][num].d==-1&&maze[x][y]!=‘*‘)
47 return true;
48 return false;
49 }
50 void bfs ( int sx,int sy,int num)
51 {
52 q[sx][sy][num].d = 0;
53 queue<int>x;
54 queue<int>y;
55 x.push(sx);
56 y.push(sy);
57 while (!x.empty())
58 {
59 int px = x.front();x.pop();
60 int py = y.front();y.pop();
61 //cout<<"px:"<<px<<" py:"<<py<<endl;
62 for ( int i = 0 ; i < 4 ; i ++ )
63 {
64 int nx = px + dx[i];
65 int ny = py + dy[i];
66 // cout<<"nx:"<<nx<<" ny:"<<ny<<endl;
67 if (ok(nx,ny,num))
68 {
69 // cout<<"why???"<<endl;
70 if (maze[nx][ny]==‘#‘)
71 {
72 // q[nx][ny][num].d = q[px][py][num].d+1;
73 q[nx][ny][num].prex = px;
74 q[nx][ny][num].prey = py;
75 // maze[nx][ny]=‘.‘;
76 }
77 else
78 {
79 // q[nx][ny][num].d = q[px][py][num].d;
80 q[nx][ny][num].prex = px;
81 q[nx][ny][num].prey = py;
82 }
83 x.push(nx);
84 y.push(ny);
85 }
86 }
87 }
88 }
89 void solve ( int x,int y)
90 {
91 // cout<<"x:"<<x<<" y:"<<y<<endl;
92 if (q[x][y][0].prey!=-1&&q[x][y][0].prex!=-1)
93 {
94 solve(q[x][y][0].prex,q[x][y][0].prey);
95 if (!v[x][y]&&maze[x][y]==‘#‘)
96 {
97 sum++;
98 // cout<<"sum:"<<sum<<endl;
99 v[x][y] = true;
100 }
101 }
102 }
103 void solve2( int x,int y)
104 {
105 // cout<<"x:"<<x<<" y:"<<y<<endl;
106 if (q[x][y][1].prex!=-1&&q[x][y][1].prey!=-1)
107 {
108 solve2(q[x][y][1].prex,q[x][y][1].prey);
109 if (!v[x][y]&&maze[x][y]==‘#‘)
110 {
111 sum++;
112 v[x][y] =true;
113 }
114 }
115 }
116 int main()
117 {
118 int T;
119 cin>>T;
120 while (T--)
121 {
122 scanf("%d %d",&h,&w);
123 int cnt = 0;
124 for ( int i = 0 ; i < h ; i ++ )
125 {
126 cin>>maze[i];
127 }
128 for ( int i = 0 ; i < h ; i ++ )
129 {
130 for ( int j = 0 ; j < w ; j ++ )
131 {
132 if (maze[i][j]==‘$‘)
133 {
134 cnt++;
135 tar[cnt][0] = i;
136 tar[cnt][1] = j;
137 // cout<<"tarx:"<<i<<" tary:"<<j<<endl;
138 }
139 }
140 }
141 cnt = 0;
142 for ( int i = 0 ; i < h ; i ++ )
143 {
144 if (maze[i][0]==‘.‘||maze[i][0]==‘#‘)
145 {
146 cnt++;
147 beg[cnt][0]=i;
148 beg[cnt][1]=0;
149 }
150 if (maze[i][w-1]==‘.‘||maze[i][w-1]==‘#‘)
151 {
152 cnt++;
153 beg[cnt][0]=i;
154 beg[cnt][1]=w-1;
155 }
156 }
157 for ( int j = 0 ; j < w ; j ++)
158 {
159 if (maze[0][j]==‘.‘||maze[0][j]==‘#‘)
160 {
161 cnt++;
162 beg[cnt][0]=0;
163 beg[cnt][1]=j;
164 }
165 if (maze[h-1][j]==‘.‘||maze[h-1][j]==‘#‘)
166 {
167 cnt++;
168 beg[cnt][0]=h-1;
169 beg[cnt][1]=j;
170 }
171 }
172 memset(q,-1,sizeof(q));
173 bfs(tar[1][0],tar[1][1],0);
174 bfs(tar[2][0],tar[2][1],1);
175 int ans = inf;
176 for ( int i = 1; i <= cnt ; i ++)
177 {
178 int tmpx = beg[i][0];
179 int tmpy = beg[i][1];
180 // if (q[tmpx][tmpy][0].d==-1||q[tmpx][tmpy][1].d==-1) continue;
181 sum = 0;
182 memset(v,false,sizeof(v));
183 solve(tmpx,tmpy);
184 solve2(tmpx,tmpy);
185 // cout<<"sum:"<<sum<<endl;
186 ans = min (ans,sum);
187 }
188 cout<<ans<<endl;
189 }
190 return 0;
191 }