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  • 【Codeforces Round #476 (Div. 2) [Thanks, Telegram!] B】Battleship

    【链接】 我是链接,点我呀:)
    【题意】

    在这里输入题意

    【题解】

    暴力枚举船的左上角。 然后统计每个点被覆盖次数就好。

    【代码】

    #include <bits/stdc++.h>
    #define LL long long
    #define rep1(i,a,b) for (int i = a;i <= b;i++)
    #define rep2(i,a,b) for (int i = a;i >= b;i--)
    #define all(x) x.begin(),x.end()
    #define pb push_back
    #define lson l,mid,rt<<1
    #define rson mid+1,r,rt<<1|1
    using namespace std;
    
    const double pi = acos(-1);
    const int dx[4] = {0,0,1,-1};
    const int dy[4] = {1,-1,0,0};
    const int N = 100;
    
    int n,k,num[N+10][N+10];
    char s[N+10][N+10];
    
    int main(){
    	#ifdef LOCAL_DEFINE
    	    freopen("rush_in.txt", "r", stdin);
    	#endif
    	ios::sync_with_stdio(0),cin.tie(0);
        cin >> n >> k;
        for (int i = 1;i <= n;i++) cin >>(s[i]+1);
        for (int i = 1;i <= n;i++)
            for (int j = 1;j <= n;j++){
                    bool ok = true;
                    for (int l = 0;l < k;l++)
                        if (j+l>n||s[i][j+l]!='.'){
                                ok = false;
                                break;
                        }
                    if (ok){
                        for (int l = 0;l < k;l++)
                            num[i][j+l]++;
                    }
    
                    ok = true;
                    for (int l = 0;l < k;l++)
                        if (i+l>n || s[i+l][j]!='.'){
                                ok = false;
                                break;
                        }
                    if (ok && k>1)
                        for (int l = 0;l < k;l++)
                            num[i+l][j]++;
    
                }
    
        int x = 1,y = 1;
        for (int i = 1;i <= n;i++)
            for (int j = 1;j <= n;j++){
                if (num[i][j]>num[x][y]){
                    x = i,y = j;
                }
            }
        cout<<x<<' '<<y<<endl;
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/8949151.html
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