首页 > 代码库 > Uva 10452 - Marcus

Uva 10452 - Marcus

 1 #include <iostream> 2 #define EOF (-1) 3  4 using namespace std; 5  6 char Path[12][100]; 7 int Step[7]={0};//记录走过的足迹 8 char example[8] = {@,I,E,H,O,V,A,#}; 9 10 void Dfs(int n,int m,int ex);11 int main(){12     int line,m,n;13     string Commands[3]={"forth","left","right"};14     //freopen("D:\\t.txt","r",stdin);15     while((cin>>line)&&(line!=EOF)){16         for(int i = 0;i < line;i++){17             cin>>n>>m;//输入鹅卵石路的宽m,长n18             for(int j=0;j<n;j++){19                 for(int k=0;k<m;k++){20                     cin>>Path[j][k];21                 }22             }23 24                 for(int q = 0;q < m;q++){25                     if(example[0] == Path[n-1][q]){26                         Dfs(n-1,q,0);27                         break;28                     }29                 }30             for(int t = 0;t < 7;t++){31                 cout<<Commands[Step[t]];32                 if(t<6)cout <<" ";33             }34             cout<<endl;35         }36     }37     return 0;38 }39 40 void Dfs(int n,int m,int ex)41 {42     if( (Path[n-1][m] == example[ex+1]) && (ex<7)) {43         Step[ex] = 0;44         Dfs(n-1,m,ex+1);45     }46     if( (Path[n][m-1] == example[ex+1]) && (ex<7)){47         Step[ex] = 1;48         Dfs(n,m-1,ex+1);49     }50     if( (Path[n][m+1] == example[ex+1]) && (ex<7)){51         Step[ex] = 2;52         Dfs(n,m+1,ex+1);53     }54 }

原来递归运算时陷入了死循环,加上“ex>7”的判定,就好了;
输入的时候注意m,n(我是新手(*^__^*) )

题目意思很简单,就是沿着一个路径走到底就行,用dfs解决。

 
 
 
 

Uva 10452 - Marcus