非常有趣的一道題目，大意是給你六種符號的16進制文本，讓你轉(zhuǎn)化成二進制并識別出來

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代碼實現(xiàn)上參考了//http://blog.csdn.net/u012139398/article/details/39533409

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#include<cstdio> #include <algorithm> #include <cstring> #include <vector> #include <set>using namespace std;const int MAXH = 210; const int MAXW = 65; int H,W; int dx[] = {0, 0, -1, 1}; int dy[] = {1, -1, 0, 0}; char* bin[16]= {"0000","0001","0010","0011","0100","0101","0110","0111" ,"1000","1001","1010","1011","1100","1101","1110","1111"}; char code[7] = "WAKJSD";char pic1[MAXH][MAXW]; char pic[MAXH][MAXW<<2]; int color[MAXH][MAXW<<2];void dfs(int x,int y,int col) {color[x][y] = col;for(int i = 0; i < 4; ++i){int nx = x + dx[i], ny = y + dy[i];if(nx >= 0 && nx < H && ny >= 0 && ny < W&& pic[x][y] == pic[nx][ny] && !color[nx][ny])dfs(nx,ny,col);} } int main() {// freopen("in.txt","r",stdin);// FEint cas = 0;while(scanf("%d%d",&H,&W),H){memset(pic,0,sizeof(pic));memset(color,0,sizeof(color));for(int i = 0; i < H; ++i)scanf("%s",pic1[i]);for(int i = 0 ; i < H; ++i)for(int j = 0; j < W; ++j){if(pic1[i][j]<='9'&&'0'<=pic1[i][j])pic1[i][j] -= '0';elsepic1[i][j] -= 'a' - 10;for(int k = 0; k < 4; ++k){//decodepic[i+1][1+(j<<2)+k] = bin[pic1[i][j]][k] - '0';}}H += 2;W = (W<<2)+2;vector<int> cc;int cnt = 0;//cnt = 1一定是白色背景for(int i = 0 ; i < H; ++i)for(int j = 0; j < W; ++j){if(!color[i][j]) {dfs(i,j,++cnt);if(pic[i][j] == 1) cc.push_back(cnt);//記錄黑色所在的連通塊 //！！沒放到括號里 }}vector< set<int> > neigh(cnt+1);for(int i = 0 ; i < H; ++i)for(int j = 0; j < W; ++j) {if(pic[i][j]){for(int k = 0; k < 4; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && x < H && y >= 0 && y < W&& pic[x][y] == 0 && color[x][y] != 1)neigh[color[i][j]].insert(color[x][y]); }}}vector<char> ans;for(int i = 0 ; i < cc.size(); ++i)ans.push_back(code[neigh[cc[i]].size()]);sort(ans.begin(),ans.end());printf("Case %d: ",++cas);for(int i = 0, sz = ans.size(); i < sz; ++i)printf("%c",ans[i]);puts("");}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/jerryRey/p/4596903.html

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