HDU-4414 Finding crosses 水题

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4414

　　直接暴力判断即可。

//STATUS:C++_AC_15MS_232KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=60;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

char ma[N][N];
int n;

int isok(int x,int y)
{
    int i,j,l,r,cnth;
    for(l=1;y-l>=0 && ma[x][y-l]=='#';l++){
        if(ma[x-1][y-l]=='#' || ma[x+1][y-l]=='#')return 0;
    }
    for(r=1;y+r<n && ma[x][y+r]=='#';r++){
        if(ma[x-1][y+r]=='#' || ma[x+1][y+r]=='#')return 0;
    }
    if(l!=r || l==1 || r==1)return 0;
    cnth=l;
    for(l=1;x-l>=0 && ma[x-l][y]=='#';l++){
        if(ma[x-l][y-1]=='#' || ma[x-l][y+1]=='#')return 0;
    }
    for(r=1;x+r<n && ma[x+r][y]=='#';r++){
        if(ma[x+r][y-1]=='#' || ma[x+r][y+1]=='#')return 0;
    }
    if(l!=r || l==1 || r==1)return 0;
    if(cnth!=l)return 0;
    return 1;
}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,cnt;
    while(~scanf("%d",&n) && n)
    {
        for(i=0;i<n;i++){
            scanf("%s",ma[i]);
        }
        cnt=0;
        for(i=1;i<n-1;i++){
            for(j=1;j<n-1;j++){
                if(ma[i][j]=='#')cnt+=isok(i,j);
            }
        }

        printf("%d
",cnt);
    }
    return 0;
}