UVA 1103 象形符号

此题是十六进制转为二进制+DFS，通过搜索图像中有几个被隔绝的白“洞”来判断是哪个文字。这里考虑一个问题，图像最外面的白色空间怎么办？想到先提前把图像外层的白色空间搜索完毕，从而搜索时只会考虑被包在中间的白色。但是如果黑色占据了边缘，隔绝了外层的白色呢？所以在这里我在整个图像的外面再加一层白色，从而使整个图像的外部白色连成一起。输出时，由于按照字典序输出，可以考虑使用vector，每次判断出一个文字时加入，最后在输出前调用一次sort()即可。
代码如下：

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<vector>#include<string>using namespace std;const int maxn = 205;char dict[16][5] = { "0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111" };char alpha[] = { 'W', 'A', 'K', 'J', 'S', 'D' };int a[maxn][maxn];int n, m;int dirx[] = { 0, 0, -1, 1 }, diry[] = { -1, 1, 0, 0 };int cur;bool isin(int r, int c){    if (r<0 || c<0 || r > n+1 || c > m+1)        return false;    return true;}void dfs1(int r, int c){//遍历白色  0    if (!isin(r, c) || a[r][c] != 0)        return;    a[r][c] = 3;    for (int i = 0; i < 4; i++)        dfs1(r + dirx[i], c + diry[i]);}void dfs2(int r, int c){    if (!isin(r, c) || a[r][c] != 1)        return;    a[r][c] = 2;    for (int i = 0; i < 4; i++){        int dr = r + dirx[i], dc = c + diry[i];        if (!isin(dr, dc))            continue;        if (a[dr][dc] == 0){            cur++;            dfs1(dr, dc);        }        else if (a[dr][dc] == 1)            dfs2(dr, dc);    }}int main(){    int kase = 0;    while (cin >> n >> m && n && m){        ++kase;        vector<char> v;        memset(a, 0, sizeof(a));        for (int i = 1; i <= n; i++){            int k = 1;            for (int j = 1; j <= m; j++){                char c;                cin >> c;                int t;                if (c >= '0'&&c <= '9')                    t = c - '0';                else                    t = c - 'a' + 10;                for (int p = 0; p < 4; p++)                    a[i][k++] = dict[t][p] - '0';            }        }        m = 4 * m;        dfs1(0, 0);        for (int i = 1; i <= n; i++){            for (int j = 1; j <= m; j++){                if (a[i][j] == 1){                    cur = 0;                    dfs2(i, j);                    v.push_back(alpha[cur]);                }            }        }        sort(v.begin(), v.end());        printf("Case %d: ", kase);        for (int i = 0; i < v.size(); i++)            cout << v[i];        cout << endl;    }    return 0;}