例题6-13 古代象形符号（Ancient Messages, World Finals 2011, UVa 1103）

思路:
1. 按照每个图形中的空白区域个数来区分图形。
2. 先进行一次dfs将外围全部填充，把区域划分为多个。
3. 对单块区域处理，从黑色开始dfs，在其中有白色则填充并计数，同时把黑色填充。

#include <iostream>#include <string>#include <vector>#include <stack>#include <queue>#include <deque>#include <set>#include <map>#include <algorithm>#include <functional>#include <utility>#include <cstring>#include <cstdio>#include <cstdlib>#include <ctime>#include <cmath>#include <cctype>#define CLEAR(a, b) memset(a, b, sizeof(a))#define IN() freopen("in.txt", "r", stdin)#define OUT() freopen("out.txt", "w", stdout)#define LL long long#define maxn 205#define maxm 6000005#define mod  10007#define INF 1000000007#define EPS 1e-7#define PI 3.1415926535898#define N 4294967296using namespace std;//-------------------------CHC------------------------------//int a[maxn][maxn];int n, m, cnt;const int dx[] = { 1, -1, 0, 0 }, dy[] = { 0, 0, 1, -1 };const char code[] = "WAKJSD";map<char, int> _hex;vector<char> ans;void dfs(int x, int y) {if (x < 0 || x > n+1 || y < 0 || y > m+1) return;a[x][y] = 2;for (int i = 0; i < 4; ++i) {int xx = x + dx[i], yy = y + dy[i];if (a[xx][yy] == 0) dfs(xx, yy);}}void dfs2(int x, int y) {if (x < 0 || x > n+1 || y < 0 || y > m+1) return;a[x][y] = 2;for (int i = 0; i < 4; ++i) {int xx = x + dx[i], yy = y + dy[i];if (a[xx][yy] == 0) dfs(xx, yy), ++cnt;else if (a[xx][yy] == 1) dfs2(xx, yy);}}void debug() {for (int i = 0; i <= n+1; ++i) {for (int j = 0; j <= m+1; ++j) printf("%d", a[i][j]);puts("");}puts("");}int main() {//IN(); OUT();int kase = 1;for (int i = 0; i < 10; ++i) _hex['0' + i] = i;for (int i = 0; i < 6; ++i) _hex['a' + i] = 10 + i;while (scanf("%d%d", &n, &m) && (n && m)) {ans.clear();CLEAR(a, 0);char s[55];for (int i = 1; i <= n; ++i) {scanf("%s", s);for (int j = 0, k = 1; j < m; ++j, k += 4) {for (int t = 0; t < 4; ++t) a[i][k + t] = _hex[s[j]] >> (3 - t) & 1;}}m <<= 2;dfs(0, 0);//debug();for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) if (a[i][j] == 1) {cnt = 0;dfs2(i, j);ans.push_back(code[cnt]);}printf("Case %d: ", kase++);sort(ans.begin(), ans.end());for (int i = 0; i < ans.size(); ++i) printf("%c", ans[i]);puts("");}return 0;}