題目描述：
Arkady is playing Battleship. The rules of this game aren’t really important.
There is a field of n×n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn’t know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
輸入描述:
The first line contains two integers n and k (1≤k≤n≤100) — the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either ‘#’ (denotes a definitely empty cell) or ‘.’ (denotes a cell that can belong to the ship).
輸出描述:
Output two integers — the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
輸入：
4 3
#…#
#.#.
…
.###
10 4
#…##…
.#…#…
…#…#…#.
…#.#…
.#…##.#…
…#…#
…#.##…
.#…#.#…
…#…#.
…#.#…#
19 6
##…###
#…#####…##
…#########…
…###########…
…#############…
…###############…
.#################.
.#################.
.#################.
.#################.
#####…##…####
####…###
####…###
#####…####…####
.#####…####…#####
…###…###…
…###########…
…##…
#…#
輸出：
3 2
6 1
1 8
題意：
有一個n×n個單元的區(qū)域與一艘長度為k寬度為1的宇宙飛船。宇宙飛船可以水平或垂直放置。但是不知道其位置。對于每個單元格，都知道它是否可以容納宇宙飛船的一部分。找到一個單元格，該單元格上宇宙飛船可能出現(xiàn)的次數(shù)最大。
題解：
枚舉每一個點開始向下向右找
代碼：

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std;const int maxn = 100 + 5; int n,k; char s[maxn][maxn]; int a[maxn][maxn];void tofind(int x,int y){int sum = 0;for(int i = x; i < min(n,x + k); i ++){if(s[i][y] == '.') sum ++;}if(sum == k){for(int i = x; i < min(n,x + k); i ++){a[i][y] ++;}}sum = 0;for(int i = y; i < min(y + k,n); i ++){if(s[x][i] == '.') sum ++;}if(sum == k){for(int i = y; i < min(y + k,n); i ++){a[x][i] ++;}} }int main(){while(scanf("%d%d",&n,&k)!=EOF){for(int i = 0; i < n; i ++) scanf("%s",s[i]);memset(a,0,sizeof(a));for(int i = 0; i < n; i ++){for(int j = 0; j < n; j ++){if(s[i][j] == '.'){tofind(i,j);}}}int maxx = 0;int x = 0,y = 0;for(int i = 0; i < n; i ++){for(int j = 0; j < n; j ++){if(a[i][j] > maxx){maxx = a[i][j];x = i;y = j;}}}printf("%d %d\n",(x + 1),(y + 1));}return 0; }

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