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  • First Step (ファーストステップ)

    老二次元题面了

    直接统计

    #include<iostream>
    using namespace std;
    
    const int N = 110;
    
    char g[N][N];
    int r, c, k;
    int p[N], q[N];
    int res;
    
    int main(){
        cin >> r >> c >> k;
        
        for(int i = 1; i <= r; i ++) cin >> g[i] + 1;
        
        for(int i = 1; i <= r; i ++){
            int cnt = 0;
            for(int j = 1; j <= c; j ++){
                while(j <= c && g[i][j] == '.') cnt ++, j ++;
                if(cnt >= k) res += cnt - k + 1;
                cnt = 0;
            }
        }
        
        for(int i = 1; i <= c; i ++){
            int cnt = 0;
            for(int j = 1; j <= r; j ++){
                while(j <= r && g[j][i] == '.') cnt ++, j ++;
                if(cnt >= k) res += cnt - k + 1;
                cnt = 0;
            }
        }
        
        if(k == 1) res /= 2;
        cout << res << endl;
        
        return 0;
    }
    

    搜索版

    #include<iostream>
    using namespace std;
    
    const int N = 110;
    
    char g[N][N];
    int r, c, k;
    int st[N][N];
    int dx[] = {1, 0}, dy[] = {0, 1};
    
    int dfs(int x, int y, int u){
        int a = x + dx[u], b = y + dy[u], cnt = 1;
        if(a <= r && b <= c && g[a][b] == '.') cnt += dfs(a, b, u);
        return cnt;
    }
    
    int main(){
        cin >> r >> c >> k;
        
        for(int i = 1; i <= r; i ++) cin >> g[i] + 1;
        
        int res = 0;
        
        for(int i = 1; i <= r; i ++)
            for(int j = 1; j <= c; j ++)
                if(g[i][j] == '.')
                    for(int u = 0; u < 2; u ++)
                        if(dfs(i, j, u) >= k) res ++;
        
        if(k == 1) res /= 2;
        cout << res;
        
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/tomori/p/13771639.html
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