2014新生暑假个人排位赛01D.BLOCKS

6 8
.....#.#
##.....#
##.....#
.......#
#......#
#..#...#
6 8
.....#.#
##.....#
###...##
.......#
##.....#
#..#...#

There are 5 ships.
So Sad

先扫一遍看有没有不联通的矩形，很简单，就是“缺角”的矩形


    
    

     
     接下来，枚举#的点,看其上方和左边是不是.即可，这样的#的数量即为ans
    
    

#include 
#include 
#include 
#define N 1005

using namespace std;

char a[N][N];
int n,m,ans;
int flag;
int judge(){
    int i=2,j=2,p,q;
    for(;i<=n;i++)
        for(;j<=m;j++){
            int tmp=0;
            for(p=i-1;p<=i;p++)
                for(q=j-1;q<=j;q++)
                    if(a[p][q]=='.')tmp++;
            if(tmp==1) return 0;
        }
    return 1;
}
int main(){
    int i,j;
    while(~scanf("%d%d",&n,&m)){
        for(i=1;i<=n;i++) scanf("%s",a[i]);
        for(i=1;i<=n;i++)
            for(j=m;j>=1;j--)
                a[i][j]=a[i][j-1];
        for(i=0;i<=max(m,n);i++) {a[0][i]='.';a[i][0]='.';}

        if(!judge()) printf("So Sad\n");
        else{
            ans=0;
            for(i=1;i<=n;i++)
                for(j=1;j<=m;j++)
                    if(a[i][j]=='#' && a[i][j-1]=='.' && a[i-1][j]=='.') ans++;
            printf("There are %d ships.\n",ans);
        }
    }
    return 0;
}