http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1119
1119: Collecting Coins
Time Limit: 3 Sec Memory Limit: 128 MB
Submit: 144 Solved: 35
[Submit][Status][Web Board]
Description
In a maze of r rows and c columns, your task is to collect as many coins as possible.
Each square is either your start point "S"(which will become empty after you leave), an empty square ".", a coin square "C" (which will become empty after you step on this square and thus collecting the coin), a rock square "O" or an obstacle square "X".
At each step, you can move one square to the up, down, left or right. You cannot leave the maze or enter an obstacle square, but you can push each rock at most once (i.e. You can treat a rock as an obstacle square after you push it).
To push a rock, you must stand next to it. You can only push the rock along the direction you‘re facing, into an neighboring empty square (you can‘t push it outside the maze, and you can‘t push it to a squarecontiaining a coin).For example, if the rock is to your immediate right, you can only push it to its right neighboring square.
Find the maximal number of coins you can collect.
Input
The first line of input contains a single integer T (T<=25), the number of test cases.
Each test case begins with two integers r and c (2<=r,c<=10), then followed by r lines, each with c columns.
There will be at most 5 rocks and at most 10 coins in each maze.
Output
For each test case, print the maximal number of coins you can collect.
Sample Input
3 3 4 S.OC ..O. .XCX 4 6 S.X.CC ..XOCC ...O.C ....XC 4 4 .SXC OO.C ..XX .CCC
Sample Output
1 6 3
HINT
Source
分析；
BFS
AC代码；
1 #include<vector>
2 #include<list>
3 #include<map>
4 #include<set>
5 #include<deque>
6 #include<stack>
7 #include<bitset>
8 #include<algorithm>
9 #include<functional>
10 #include<numeric>
11 #include<utility>
12 #include<sstream>
13 #include<iostream>
14 #include<iomanip>
15 #include<cstdio>
16 #include<cmath>
17 #include<cstdlib>
18 #include<cstring>
19 #include<ctime>
20 #define LL long long
21
22 using namespace std;
23 int mp[20][20];
24 int vis[20][20];
25 int xadd[] = {1,-1,0,0};
26 int yadd[] = {0,0,1,-1};
27 struct node
28 {
29 int x;int y;
30 int is;
31 node(int _x, int _y, int _is)
32 {
33 x = _x;
34 y = _y;
35 is = 0;
36 }
37 };
38 vector<node> C;
39 int mx = 0;
40 int ansnum = 0 ;
41 int n , m ;
42 void debug()
43 {
44 for(int i = 1;i <= n;i ++)
45 {
46 for(int j = 1;j <= m;j ++)
47 printf("%d ",mp[i][j]);
48 printf("\n");
49 }
50 }
51 void bfs(int x, int y ,int ans)
52 {
53 //printf("%d %d\n",x,y);
54 if(mx == ansnum)
55 return;
56 vector<node> Q;
57 Q.push_back(node(x,y,0));
58 vis[x][y] = 1;
59 int l = 0;
60 int r = 0;
61 while(l <= r )
62 {
63 for(int i = 0 ;i <= 3;i ++)
64 {
65 int tx = Q[l].x + xadd[i] ;
66 int ty = Q[l].y + yadd[i] ;
67 if(mp[tx][ty] >= 1 && !vis[tx][ty])
68 {
69 vis[tx][ty] = 1;
70 r ++ ;
71 if(mp[tx][ty] == 2)
72 {
73 Q.push_back(node(tx,ty,1));
74 ans ++ ;
75 }
76 else Q.push_back(node(tx,ty,0));
77 }
78 }
79 l ++ ;
80 }
81 if(ans > mx)
82 mx = ans;
83 for(int i = 0;i < C.size();i ++)
84 {
85 if(!C[i].is)
86 {
87 for(int s = 0 ;s <= 3;s ++)
88 {
89 int tx = C[i].x + xadd[s];
90 int ty = C[i].y + yadd[s];
91 int ttx = C[i].x - xadd[s];
92 int tty = C[i].y - yadd[s];
93 //printf("%d %d %d %d\n",tx,ty,ttx,tty);
94 if(mp[tx][ty] == 1 && vis[ttx][tty] == 1)
95 {
96 mp[tx][ty] = -1;
97 mp[C[i].x][C[i].y] = 1;
98 C[i].is = 1;
99 bfs(C[i].x,C[i].y,ans);
100 mp[tx][ty] = 1;
101 mp[C[i].x][C[i].y] = 0;
102 C[i].is = 0 ;
103 }
104 }
105 }
106 }
107 for(int i = r; i >= 0 ;i --)
108 {
109 vis[Q[i].x][Q[i].y] = 0 ;
110 if(Q[i].is)
111 {
112 mp[Q[i].x][Q[i].y] = 2 ;
113 }
114 }
115 }
116 int main(){
117 int t ;
118 scanf("%d",&t);
119 while(t--)
120 {
121 memset(mp,-1,sizeof(mp));
122 memset(vis,0,sizeof(vis));
123 scanf("%d %d",&n,&m);
124 char str[20];
125 int bex, bey ;
126 ansnum =0 ;
127 C.clear();
128 for(int i = 1 ;i <= n;i ++)
129 {
130 scanf("%s",&str[1]);
131 for(int j = 1;j <= m; j ++)
132 {
133 if(str[j] == ‘S‘)
134 {
135 mp[i][j] = 1 ;
136 bex = i ;
137 bey = j ;
138 }else if(str[j] == ‘C‘)
139 {
140 ansnum ++;
141 mp[i][j] = 2;
142 }else if(str[j] == ‘X‘)
143 {
144 mp[i][j] = -1;
145 }else if (str[j] == ‘O‘){
146 mp[i][j] = 0 ;
147 C.push_back(node(i,j,0));
148 }else {
149 mp[i][j] = 1;
150 }
151 }
152 }
153 mx = 0 ;
154 bfs(bex,bey,0);
155 printf("%d\n",mx);
156 }
157
158 return 0;
159 }