题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5612
题意：给出一个矩阵，在（i*2，j*2）（i，j = 0, 1, 2...）是数字0-9，数字之间是+，-，*，/运算符。其他位置是#号。
问给出一个数字，有没有可能在矩阵中找到一条路线可以计算得到这个数字。
思路：DFS搜索
1 #include <bits/stdc++.h> 2 using namespace std; 3 char mp[35][35]; 4 int T; 5 int vis[35][35]; 6 int n, m; 7 double sum; 8 #define eps 1e-10 9 bool flag; 10 void dfs(int x, int y, double val) 11 { 12 if(flag) return; 13 if(fabs(sum-val) < eps) 14 { 15 flag = true; return; 16 } 17 if(x-2 >= 0) 18 { 19 if(!vis[x-2][y]) 20 { 21 vis[x-2][y] = 1; 22 if(mp[x-1][y] == '*') dfs(x-2, y, val*(mp[x-2][y]-'0')); 23 else if(mp[x-1][y] == '+') dfs(x-2, y, val+(mp[x-2][y]-'0')); 24 else if(mp[x-1][y] == '-') dfs(x-2, y, val-(mp[x-2][y]-'0')); 25 else if(mp[x-1][y] == '/') dfs(x-2, y, val/(mp[x-2][y]-'0')); 26 vis[x-2][y] = 0; 27 } 28 } 29 if(x+2 <= n-1) 30 { 31 if(!vis[x+2][y]) 32 { 33 vis[x+2][y] = 1; 34 if(mp[x+1][y] == '*') dfs(x+2, y, val*(mp[x+2][y]-'0')); 35 else if(mp[x+1][y] == '+') dfs(x+2, y, val+(mp[x+2][y]-'0')); 36 else if(mp[x+1][y] == '-') dfs(x+2, y, val-(mp[x+2][y]-'0')); 37 else if(mp[x+1][y] == '/') dfs(x+2, y, val/(mp[x+2][y]-'0')); 38 vis[x+2][y] = 0; 39 40 } 41 } 42 if(y-2 >= 0) 43 { 44 if(!vis[x][y-2]) 45 { 46 vis[x][y-2] = 1; 47 if(mp[x][y-1] == '*') dfs(x, y-2, val*(mp[x][y-2]-'0')); 48 else if(mp[x][y-1] == '+') dfs(x, y-2, val+(mp[x][y-2]-'0')); 49 else if(mp[x][y-1] == '-') dfs(x, y-2, val-(mp[x][y-2]-'0')); 50 else if(mp[x][y-1] == '/') dfs(x, y-2, val/(mp[x][y-2]-'0')); 51 vis[x][y-2] = 0; 52 } 53 } 54 55 if(y+2 <= m-1) 56 { 57 if(!vis[x][y+2]) 58 { 59 vis[x][y+2] = 1; 60 if(mp[x][y+1] == '*') dfs(x, y+2, val*(mp[x][y+2]-'0')); 61 else if(mp[x][y+1] == '+') dfs(x, y+2, val+(mp[x][y+2]-'0')); 62 else if(mp[x][y+1] == '-') dfs(x, y+2, val-(mp[x][y+2]-'0')); 63 else if(mp[x][y+1] == '/') dfs(x, y+2, val/(mp[x][y+2]-'0')); 64 vis[x][y+2] = 0; 65 } 66 } 67 68 return; 69 70 } 71 int main() 72 { 73 scanf("%d", &T); 74 while(T--) 75 { 76 scanf("%d%d%lf", &n, &m, &sum); 77 for(int i = 0; i < n; i++) 78 { 79 scanf("%s", mp[i]); 80 } 81 82 flag = false; 83 for(int i = 0; i < n; i+=2) 84 { 85 for(int j = 0; j < m; j+=2) 86 { 87 memset(vis, 0, sizeof(vis)); 88 vis[i][j] = 1; 89 dfs(i, j, mp[i][j]-'0'); 90 if(flag == true) break; 91 } 92 if(flag == true) break; 93 } 94 if(flag) printf("Possible "); 95 else printf("Impossible "); 96 } 97 return 0; 98 }
