BFS——BMW图形文字识别

有一些由“_"和"#"组成的图形，图形中由"#"拼成了"B"、"M"、"W"三种字母，。用程序识别这些字母，统计三个字母的个数。

___###############________________________________

___###################____________________________

___#####################__________________________
___#####_____________#####________________________
___#####_______________###________________________
___#####_______________###________________________
___#####_______________###________________________
___#####_____________#####________________________
___#####___________#####__________________________
___###################____________________________
___###################____________________________
___#####___________#######________________________
___#####_______________###________________________
___#####_______________#####______________________
___#####_______________#####______________________
___#####_______________#####______________________
___#####_______________#####______________________
___#####_____________#####________________________
___#######################________________________
___###################____________________________

1 0 0

先BFS判断联通块

在判断首行间隔数目

#include<stdio.h>
#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
 
int hash1[209][209];
int map[209][209];
int ne[209][209];
int diri[4]={0,-1,0,1};
int dirj[4]={-1,0,1,0};
 
int used[10099];
 
struct data
{
    int ni,nj;
};
 
int n,m;
 
void bfs(int x,int y,int add)
{
    queue<data>qq;
    int i,j;
    data f,s,t;
    f.ni=x;
    f.nj=y;
    qq.push(f);
    while(!qq.empty())
    {
        s=qq.front();
        qq.pop();
        for(i=0;i<=3;i++)
        {
            t.ni=s.ni+diri[i];
            t.nj=s.nj+dirj[i];
             
            if(t.ni>=1&&t.ni<=n&&t.nj>=1&&t.nj<=m&&(hash1[t.ni][t.nj]==0)&&map[t.ni][t.nj]==1)
            {
                hash1[t.ni][t.nj]=add;
                qq.push(t);
            }
        }
    }
}
 
int main()
{
    char temp;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(hash1,0,sizeof(hash1));
        memset(used,0,sizeof(used));
         
        int i,j;
        char ss[1009];
        for(i=1;i<=n;i++)
        {
            //getchar();
            scanf("%s",ss);
            for(j=1;j<=m;j++)
            {
                temp=ss[j-1];
                if(temp=='_')
                    map[i][j]=0;
                else
                    map[i][j]=1;
            }
        }
         
         
        int add=1;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(hash1[i][j]==0&&map[i][j]==1)
                {
                    bfs(i,j,add);
                    add++;
                }
            }
        }
        /*
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                printf("%d",hash1[i][j]);
            }
            printf("\n");
        }*/
         
        add=1;
        int k;
        int B=0,M=0,W=0;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(hash1[i][j]==0)continue;
                if(used[hash1[i][j]]==1)continue;
                 
                add==hash1[i][j];
                 
                int temp=1;
                for(k=j+1;k<=m;k++)
                {
                    if(hash1[i][k]!=0&&hash1[i][k]!=add)
                    {
                        break;
                    }
                     
                    if(hash1[i][k]!=0&&used[hash1[i][k]]==0)
                    {
                        if(hash1[i][k-1]==0)
                            temp++;
                    }
                     
                }
                j=k;
                 
                used[add]=1;
                if(temp==1)
                {
                //  printf("%d %d\n",i,j);
                    B++;
                }
                if(temp==2)
                    M++;
                if(temp==3)
                    W++;
                add++;
            }
        }
         
        printf("%d %d %d\n",B,M,W);
         
    }
    return 0;
}