matrix_world_final_2011

C  http://acm.hust.edu.cn/vjudge/contest/view.action?cid=98613#problem/C

题意：输入16进制的n*m矩阵，其在二进制表示下有6种图中的哪几种。

解法：6种图黑点都是连通的，所以第一步先把黑的连通块标号，第二步6种图中白连通块的个数正好是0-5个，第二步每找到一个白块就找到它的边界，对应的黑块+1，最后每一个黑块中有多少白块就统计完了，按字典序输出即可。

 //#define debug
 //#define txtout
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cmath>
 #include<cctype>
 #include<ctime>
 #include<iostream>
 #include<algorithm>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #define mt(a,b) memset(a,b,sizeof(a))
 using namespace std;
 typedef long long LL;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int inf=0x3f3f3f3f;
 const int M=2e2+;
 char a[M][M];
 char str[]="WAKJSD";
 bool black[M][M];
 int index_of_black;
 int id[M][M];
 int dx[]={,,,-};
 int dy[]={,-,,};
 int n,m;
 int sum[];
 int count_of_black[M*M];
 int id_of_black;
 vector<char> answer;
 int to_int(char c){
     if(isdigit(c)) return c-'';
     return c-'a'+;
 }
 void init_black(){
     for(int i=;i<n;i++){
         int len=;
         for(int j=;j<m;j++){
             int value=to_int(a[i][j]);
             for(int k=;k>=;k--){
                 black[i][len++]=(value>>k)&;
             }
         }
     }
     m<<=;
 }
 void init_id(){
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             id[i][j]=;
         }
     }
 }
 void init_count_of_black(){
     for(int i=;i<=index_of_black;i++){
         count_of_black[i]=;
     }
 }
 bool inside(int x,int y){
     return x>=&&x<n&&y>=&&y<m;
 }
 void dfs_black(int x,int y,int Index){
     id[x][y]=Index;
     for(int i=;i<;i++){
         int tx=x+dx[i];
         int ty=y+dy[i];
         if(!inside(tx,ty)) continue;
         if(!black[tx][ty]) continue;
         if(id[tx][ty]) continue;
         dfs_black(tx,ty,Index);
     }
 }
 void solve_black(){
     index_of_black=;
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             if(!black[i][j]) continue;
             if(id[i][j]) continue;
             dfs_black(i,j,++index_of_black);
         }
     }
 }
 void dfs_white(int x,int y){
     id[x][y]=-;
     for(int i=;i<;i++){
         int tx=x+dx[i];
         int ty=y+dy[i];
         if(!inside(tx,ty)){
             id_of_black=;
             continue;
         }
         if(id[tx][ty]){
             if(id[tx][ty]>&&id_of_black==-){
                 id_of_black=id[tx][ty];
             }
             continue;
         }
         dfs_white(tx,ty);
     }
 }
 void solve_white(){
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             if(id[i][j]) continue;
             id_of_black=-;
             dfs_white(i,j);
             if(id_of_black==) continue;
             count_of_black[id_of_black]++;
         }
     }
 }
 void solve(){
     init_black();
     init_id();
     solve_black();
     init_count_of_black();
     solve_white();
     mt(sum,);
     for(int i=;i<=index_of_black;i++){
         sum[count_of_black[i]]++;
     }
     answer.clear();
     for(int i=;i<;i++){
         for(int j=;j<sum[i];j++){
             answer.push_back(str[i]);
         }
     }
     sort(answer.begin(),answer.end());
 }
 int main(){
     #ifdef txtout
     freopen("in.txt","r",stdin);
     freopen("out.txt","w",stdout);
     #endif
     int cas=;
     while(~scanf("%d%d",&n,&m),n|m){
         for(int i=;i<n;i++){
             scanf("%s",a[i]);
         }
         solve();
         printf("Case %d: ",cas++);
         int len=answer.size();
         for(int i=;i<len;i++){
             putchar(answer[i]);
         }
         puts("");
     }
     return ;
 }

K http://acm.hust.edu.cn/vjudge/contest/view.action?cid=98613#problem/K

题意：求最小的能让任意多边形垂直下落通过的间距。

解法：先做凸包，求凸包的宽度即是答案。

 //#define debug
 //#define txtout
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cmath>
 #include<cctype>
 #include<ctime>
 #include<iostream>
 #include<algorithm>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #define mt(a,b) memset(a,b,sizeof(a))
 using namespace std;
 typedef long long LL;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int inf=0x3f3f3f3f;
 const int M=1e2+;
 struct point {
     double x,y;
 } p[M],res[M];
 int n;
 char buffer[M];
 class Convex_Hull { ///凸包
     bool mult(point sp,point ep,point op) { ///>包括凸包边上的点,>=不包括。
         return (sp.x-op.x)*(ep.y-op.y)>=(ep.x-op.x)*(sp.y-op.y);
     }
     static bool cmp(const point &a,const point &b) { ///整数不需fabs，a.y==b.y。
         return a.y<b.y||(fabs(a.y-b.y)<eps&&a.x<b.x);
     }
 public:
     int Graham(int n,point p[],point res[]) { ///传入点的个数，点数组p[],凸包点存在res[],返回凸包点数.
         sort(p,p+n,cmp);
         if(n==) return ;
         res[]=p[];
         if(n==) return ;
         res[]=p[];
         if(n==) return ;
         res[]=p[];
         int top=;
         for(int i=; i<n; i++) {
             while(top&&mult(p[i],res[top],res[top-])) {
                 top--;
             }
             res[++top]=p[i];
         }
         int len=top;
         res[++top]=p[n-];
         for(int i=n-; i>=; i--) {
             while(top!=len&&mult(p[i],res[top],res[top-])) {
                 top--;
             }
             res[++top]=p[i];
         }
         return top;
     }
 } gx;
 class Convex_Hull_Wide{///凸包的宽
     double xmult(point p1,point p2,point p0) { ///计算向量叉积(P1-P0)x(P2-P0)
         return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
     }
     double Square(double x) { ///平方
         return x*x;
     }
     double Distance(point a,point b) { ///平面两点距离
         return sqrt(Square(a.x-b.x)+Square(a.y-b.y));
     }
     double get(point a,point b,point c){
         return fabs(xmult(a,b,c));
     }
 public:
     double Rotate_Calipers_Wide(point p[],int n) {///传入凸包和点数
         double result=-;
         for(int i=,j=; i<n; i++) {
             while(get(p[j+],p[i+],p[i])>get(p[j],p[i+],p[i])){
                 j=(j+)%n;
             }
             double temp=get(p[j],p[i+],p[i])/Distance(p[i],p[i+]);
             if(result==-){
                 result=temp;
                 continue;
             }
             result=min(result,temp);
         }
         return result;
     }
 }ts;
 void round_up(double &x){
     sprintf(buffer,"%.6f",x);
     int len=strlen(buffer);
     for(int i=;i<len;i++){
         if(buffer[i]=='.'){
             if(buffer[i+]!=''){
                 buffer[i+]='';
             }
             break;
         }
     }
     sscanf(buffer,"%lf",&x);
 }
 double solve() {
     int len=gx.Graham(n,p,res);
     double result=ts.Rotate_Calipers_Wide(res,len);
     round_up(result);
     return result;
 }
 int main() {
 #ifdef txtout
     freopen("in.txt","r",stdin);
     freopen("out.txt","w",stdout);
 #endif
     int cas=;
     while(~scanf("%d",&n),n) {
         for(int i=; i<n; i++) {
             scanf("%lf%lf",&p[i].x,&p[i].y);
         }
         printf("Case %d: %.2f\n",cas++,solve());
     }
     return ;
 }

end