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-8;
const double pi=acos(-1.0);
const int inf=0x3f3f3f3f;
const int M=2e2+10;
char a[M][M];
char str[8]="WAKJSD";
bool black[M][M];
int index_of_black;
int id[M][M];
int dx[8]={0,0,1,-1};
int dy[8]={1,-1,0,0};
int n,m;
int sum[8];
int count_of_black[M*M];
int id_of_black;
vector<char> answer;
int to_int(char c){
    if(isdigit(c)) return c-'0';
    return c-'a'+10;
}
void init_black(){
    for(int i=0;i<n;i++){
        int len=0;
        for(int j=0;j<m;j++){
            int value=to_int(a[i][j]);
            for(int k=3;k>=0;k--){
                black[i][len++]=(value>>k)&1;
            }
        }
    }
    m<<=2;
}
void init_id(){
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            id[i][j]=0;
        }
    }
}
void init_count_of_black(){
    for(int i=1;i<=index_of_black;i++){
        count_of_black[i]=0;
    }
}
bool inside(int x,int y){
    return x>=0&&x<n&&y>=0&&y<m;
}
void dfs_black(int x,int y,int Index){
    id[x][y]=Index;
    for(int i=0;i<4;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=0;
    for(int i=0;i<n;i++){
        for(int j=0;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]=-1;
    for(int i=0;i<4;i++){
        int tx=x+dx[i];
        int ty=y+dy[i];
        if(!inside(tx,ty)){
            id_of_black=0;
            continue;
        }
        if(id[tx][ty]){
            if(id[tx][ty]>0&&id_of_black==-1){
                id_of_black=id[tx][ty];
            }
            continue;
        }
        dfs_white(tx,ty);
    }
}
void solve_white(){
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            if(id[i][j]) continue;
            id_of_black=-1;
            dfs_white(i,j);
            if(id_of_black==0) continue;
            count_of_black[id_of_black]++;
        }
    }
}
void solve(){
    init_black();
    init_id();
    solve_black();
    init_count_of_black();
    solve_white();
    mt(sum,0);
    for(int i=1;i<=index_of_black;i++){
        sum[count_of_black[i]]++;
    }
    answer.clear();
    for(int i=0;i<6;i++){
        for(int j=0;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=1;
    while(~scanf("%d%d",&n,&m),n|m){
        for(int i=0;i<n;i++){
            scanf("%s",a[i]);
        }
        solve();
        printf("Case %d: ",cas++);
        int len=answer.size();
        for(int i=0;i<len;i++){
            putchar(answer[i]);
        }
        puts("");
    }
    return 0;
}

 

 

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-8;
const double pi=acos(-1.0);
const int inf=0x3f3f3f3f;
const int M=1e2+10;
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==0) return 0;
        res[0]=p[0];
        if(n==1) return 1;
        res[1]=p[1];
        if(n==2) return 2;
        res[2]=p[2];
        int top=1;
        for(int i=2; i<n; i++) {
            while(top&&mult(p[i],res[top],res[top-1])) {
                top--;
            }
            res[++top]=p[i];
        }
        int len=top;
        res[++top]=p[n-2];
        for(int i=n-3; i>=0; i--) {
            while(top!=len&&mult(p[i],res[top],res[top-1])) {
                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=-1;
        for(int i=0,j=1; i<n; i++) {
            while(get(p[j+1],p[i+1],p[i])>get(p[j],p[i+1],p[i])){
                j=(j+1)%n;
            }
            double temp=get(p[j],p[i+1],p[i])/Distance(p[i],p[i+1]);
            if(result==-1){
                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=0;i<len;i++){
        if(buffer[i]=='.'){
            if(buffer[i+3]!='0'){
                buffer[i+3]='9';
            }
            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=1;
    while(~scanf("%d",&n),n) {
        for(int i=0; i<n; i++) {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        printf("Case %d: %.2f\n",cas++,solve());
    }
    return 0;
}

 

 

end