Educational Codeforces Round 25 Five-In-a-Row(DFS)

题目网址：http://codeforces.com/contest/825/problem/B

题目：

 

Alice and Bob play 5-in-a-row game. They have a playing field of size 10 × 10. In turns they put either crosses or noughts, one at a time. Alice puts crosses and Bob puts noughts.

In current match they have made some turns and now it's Alice's turn. She wonders if she can put cross in such empty cell that she wins immediately.

Alice wins if some crosses in the field form line of length not smaller than 5. This line can be horizontal, vertical and diagonal.

Input

You are given matrix 10 × 10 (10 lines of 10 characters each) with capital Latin letters 'X' being a cross, letters 'O' being a nought and '.' being an empty cell. The number of 'X' cells is equal to the number of 'O' cells and there is at least one of each type. There is at least one empty cell.

It is guaranteed that in the current arrangement nobody has still won.

Output

Print 'YES' if it's possible for Alice to win in one turn by putting cross in some empty cell. Otherwise print 'NO'.

Examples

input

XX.XX.....
.....OOOO.
..........
..........
..........
..........
..........
..........
..........
..........

output

YES

input

XXOXX.....
OO.O......
..........
..........
..........
..........
..........
..........
..........
..........

output

NO

思路：因为只有10*10，所以直接用了暴力搜索。
代码：

 #include <cstdio>
 #include <vector>
 using namespace std;
 struct node{
     int x,y;
 }dir[]={{,},{,-},{,},{-,},{-,-},{-,},{,},{,-}};
 int row[];
 int col[];
 int ok=;
 char graph[][];
 vector<node>v;
 bool check(int x,int y){
     if(x< || x>=)    return false;
     if(y< || y>=)    return false;
     if(graph[x][y]!='X')    return false;
     return true;
 }
 void dfs(int x,int y,int d,int num){
     if(!check(x, y))    return ;
     if(num==){
         ok=;
         return ;
     }
     int xt=x+dir[d].x;
     int yt=y+dir[d].y;
     dfs(xt,yt,d,num+);

 }
 int main(){
     for (int i=; i<; i++) {
         gets(graph[i]);
         for(int j=;j<;j++){
             if(graph[i][j]=='X'){
                 node t;
                 t.x=i;
                 t.y=j;
                 v.push_back(t);
             }
         }
     }
     for (int i=; i< && !ok; i++) {
         for (int j=; j< && !ok; j++) {
             if(graph[i][j]!='.')    continue;
             graph[i][j]='X';
             for(int i=;i<v.size() && !ok;i++){
                 for(int d=;d<;d++)    dfs(v[i].x,v[i].y,d,);
             }
             graph[i][j]='.';
         }
     }
     if(ok)  printf("YES\n");
     else printf("NO\n");
     return ;
 }