UVA - 1103 Ancient Messages
题意：给定一个图像，字典序输出里面对应的象形文字。
分析：把由十六进制构成的图转换成二进制的图，就能很容易理解题意。
1、在最外圈加一圈0，便于使所有象形文字外的0都是连通的
2、将所有连通的0和1都标号（dfs）
3、再查象形文字里的洞的个数，把所有1附近的不是外圈的0都set一下就好了。
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) typedef long long ll; typedef unsigned long long llu; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const ll LL_INF = 0x3f3f3f3f3f3f3f3f; const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const double eps = 1e-8; const int MAXN = 200 + 10; const int MAXT = 50 + 10; using namespace std; char pic[MAXN][MAXT << 2]; int vis[MAXN][MAXT << 2]; char s[MAXN]; map<char, string> mp; map<int, char> ma; set<int> in[44100];//连通块1里的洞的个数,set去重 vector<int> y;//由1组成的连通块的标号 vector<char> ans; int cnt, H, W;; void init(){//十六进制与二进制映射 string a = "0123456789abcdef"; string b = "0000000100100011010001010110011110001001101010111100110111101111"; int len = a.size(); int st = 0; for(int i = 0; i < len; ++i){ mp[a[i]] = b.substr(st, 4);//从st开始取4个 st += 4; } string tmp = "WAKJSD"; for(int i = 0; i < 6; ++i){ ma[i] = tmp[i]; } } bool judge(int x, int y){ return x >= 0 && x <= H + 1 && y >= 0 && y <= 4 * W + 1; } void dfs(int x, int y){ vis[x][y] = cnt; for(int i = 0; i < 4; ++i){//题意中other black pixel on its top, bottom, left, or right side int tmpx = x + dr[i]; int tmpy = y + dc[i]; if(judge(tmpx, tmpy) && !vis[tmpx][tmpy] && pic[tmpx][tmpy] == pic[x][y]){ dfs(tmpx, tmpy); } } } int main(){ init(); int cases = 0; while(scanf("%d%d", &H, &W) == 2){ if(!H && !W) return 0; memset(pic, '0', sizeof pic); memset(vis, 0, sizeof vis); for(int i = 0; i < 44100; ++i){ in[i].clear(); } y.clear(); ans.clear(); for(int i = 1; i <= H; ++i){ scanf("%s", s); string ans = ""; for(int j = 0; j < W; ++j){ ans += mp[s[j]]; } for(int j = 1; j <= 4 * W; ++j){ pic[i][j] = ans[j - 1]; } } cnt = 0;//为每一个连通块标号 for(int i = 0; i < H + 2; ++i){ for(int j = 0; j < 4 * W + 2; ++j){ if(!vis[i][j]){ ++cnt; dfs(i, j); if(pic[i][j] == '1') y.push_back(cnt); } } } for(int i = 0; i < H + 2; ++i){ for(int j = 0; j < 4 * W + 2; ++j){ if(pic[i][j] == '1'){ for(int k = 0; k < 4; ++k){ int tmpx = i + dr[k]; int tmpy = j + dc[k]; if(judge(tmpx, tmpy) && pic[tmpx][tmpy] == '0' && vis[tmpx][tmpy] != 1){//象形文字里的0 in[vis[i][j]].insert(vis[tmpx][tmpy]); } } } } } int len = y.size(); for(int i = 0; i < len; ++i){ ans.push_back(ma[in[y[i]].size()]); } sort(ans.begin(), ans.end()); len = ans.size(); printf("Case %d: ", ++cases); for(int i = 0; i < len; ++i){ printf("%c", ans[i]); } printf("\n"); } return 0; }
一组完整的测试样例：
100 25 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 00000f8000000000000000000 00001fe000000000000000000 00007ff000000000000000000 00007ff800000000000000000 0000f8f800000000000000000 0001f07c00000000000000000 0001e03c00000000001800000 0001e01c00000000003c00000 0001c01c00000000007c00000 0003c01e0000000000f800000 0003c01e0000000001f000000 0001c01c0000000003f000000 0001c01c0000000007e000000 0001e01c000000000fc000000 0001e03c000000001fc000000 0000e03c000000001fc000000 0000f038000000003ff000000 0000f078000000003ff800000 00007870000000007ff800000 000038f0000000007cfc00000 00003ce0000000007c7c00000 00781fc0f0000000f87c00000 007ffffff0000000f07c00000 007ffffff0000000f07c00000 007ffffff0000001f07c00000 007ffffff0000000e03e00000 007fcf81f0000000603e00000 00000f8000000000003e00000 00000f8000000000003e00000 00000f8000000000003e00000 00000f8000000000001e00000 00000f8000000000001f00000 00000fc000000000001f00000 00000fc000000000001f00000 00000fc000000000001f00000 00000fc000000000000f00000 00001fc000000000000f80000 00001fc000000000000f80000 00001fc000000000000f80000 00001fc000000000000f80000 00001fe000000000000f80000 00001fe000000000000780000 00001fe0000000000007c0000 00001fe0000000000007c0000 00003fe0000000000007c0000 00003fe0000000000007c0000 00003fe0000000000007c0000 00003fe0000c00000003c0000 00000000003ff0000003c0000 00000000007ff8000003e0000 0000000001fffc000003e0000 0000000003e03f000003e0000 0000000007c00f000003e0000 000000000f0003800003f0000 000000000e0001c00003fc000 000000001c0001e00007fe000 000000003c0000e0000fff000 000000073c000070000fdf000 0000001ff8000070001f0f800 0000001ff8000070001e07800 0000003cf0000078001e03800 0000003870000033001e03800 000000307800003fc01e03800 000000703800007fe00e03800 000000703800007ce00e03800 000000703c000078700703800 000000701e0000f0700701000 000000701e0000e0700300000 000000700f0001c0700000000 0000006007800380600000000 000000e003e00700600000000 000000e001fe7e00600000000 000000e000fffc00e00000000 000000e0000ff000e00000000 000000f800038000e00000000 000000fff0000000e00000000 000000fffff00000e00000000 00000003ffffe000c00000000 0000000007ffffc0c00000000 000000000007ffffc00000000 0000000000000fffc00000000 000000000000001fc00000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000 0000000000000000000000000
输出：AKW
