UVa 1419 - Ugly Windows (DFS + 暴力)

题意

找出有几个在顶端的窗口。需要满足以下条件
窗口里面只有'.'
窗口边缘是完整的。

思路

在浏览器标签上呆了一个星期的题目=。= 

今天标签实在是放不下了，就去A这题，空一点位置╮(╯▽╰)╭
找出完整的窗口。
判断这个窗口里面是不是全部是.

如何找出完整的窗口？ 

碰到一个字母就DFS，如果从下往上走到起点，说明这个窗口是完整的。因此在起点的时候不能往下走。

如何判断2？ 

枚举出当前窗口的边界，for一下。

代码

#include <cstdio>

#include <stack>

#include <set>

#include <iostream>

#include <string>

#include <vector>

#include <queue>

#include <functional>

#include <cstring>

#include <algorithm>

#include <cctype>

#include <string>

#include <map>

#include <cmath>

#define LL long long

#define ULL unsigned long long

#define SZ(x) (int)x.size()

#define Lowbit(x) ((x) & (-x))

#define MP(a, b) make_pair(a, b)

#define MS(arr, num) memset(arr, num, sizeof(arr))

#define PB push_back

#define F first

#define S second

#define ROP freopen("input.txt", "r", stdin);

#define MID(a, b) (a + ((b - a) >> 1))

#define LC rt << 1, l, mid

#define RC rt << 1|1, mid + 1, r

#define LRT rt << 1

#define RRT rt << 1|1

#define BitCount(x) __builtin_popcount(x)

#define BitCountll(x) __builtin_popcountll(x)

#define LeftPos(x) 32 - __builtin_clz(x) - 1

#define LeftPosll(x) 64 - __builtin_clzll(x) - 1

const double PI = acos(-1.0);

const int INF = 0x3f3f3f3f;

using namespace std;

const double eps = 1e-8;

const int MAXN = 100 + 10;

const int MOD = 1000007;

const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };  //0123，上下左右

typedef pair<int, int> pii;

typedef vector<int>::iterator viti;

typedef vector<pii>::iterator vitii;

char mp[MAXN][MAXN];

int vis[MAXN][MAXN], n, m;

pii tar;

vector<char> ans;

bool Check(char curWindow)

{

int i, j;

for (i = tar.F, j = tar.S; mp[i][j] == curWindow; j++);

int len = j - 2;

for (i = tar.F, j = tar.S; mp[i][j] == curWindow; i++);

int wide = i - 2;

for (i = tar.F + 1; i <= wide; i++)

for (j = tar.S + 1; j <= len; j++)

if (mp[i][j] != '.') return false;

return true;

}

void DFS(int x, int y, char curWindow)

{

if (mp[x][y] == '.' || mp[x][y] != curWindow) return;

vis[x][y] = 1;

for (int i = 0; i < 4; i++)

{

if (i == 1 && x == tar.F && y == tar.S) continue;   //一开始的时候不能往下走

int xx = x + dir[i][0], yy = y + dir[i][1];

if (i == 0 && xx == tar.F && yy == tar.S)   //如果往上走能到达起点

{

if (Check(curWindow))

ans.PB(curWindow);

}

if (xx >= 1 && xx <= n && yy >= 1 && yy <= m && !vis[xx][yy])

DFS(xx, yy, curWindow);

}

}

int main()

{

//ROP;

int i, j;

while (scanf("%d%d", &n, &m), n + m)

{

ans.clear();

MS(vis, 0);

for (i = 1; i <= n; i++) scanf("%s", mp[i] + 1);

for (i = 1; i <= n; i++)

for (j = 1; j <= m; j++)

if (!vis[i][j] && mp[i][j] != '.')

{

tar.F = i, tar.S = j;

DFS(i, j, mp[i][j]);

}

sort(ans.begin(), ans.end());

for (i = 0; i < SZ(ans); i++) putchar(ans[i]);

puts("");

}

return 0;

}