2017微软秋季校园招聘在线编程笔试

//寻找M S的数量 2017微软秋季校园招聘在线编程笔试 hihocoder 1402 MS Recognition
//看了讨论区写了一发
//套一下思路：
//有一个很简单的做法，一遍bfs找到一个端点，再一遍bfs找到另一个端点，然后把两个端点的中点和整个图形的重心比对一下，区分度非常大。具体来说就是比对一下端点中点到重心的距离，以及两个端点的距离的一半，二者相除，判断是否超过0.5即可。
#include <bits/stdc++.h>
using namespace std;
#define LL long long
typedef pair<int,int> pii;
const int inf = 0x3f3f3f3f;
const int N =1e5+10;
#define clc(a,b) memset(a,b,sizeof(a))
const double eps = 1e-8;
const int MOD = 1e9+7;
void fre() {freopen("in.txt","r",stdin);}
void freout() {freopen("out.txt","w",stdout);}
inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}

int sgn(double x) {
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}

struct Point {
    int x,y;
    Point() {}
    Point(int _x,int _y) {
        x = _x;
        y = _y;
    }
    Point operator -(const Point &b)const {
        return Point(x - b.x,y - b.y);
    }
    int operator ^(const Point &b)const {
        return x*b.y - y*b.x;
    }
    int operator *(const Point &b)const {
        return x*b.x + y*b.y;
    }
    friend int dis2(Point a) {
        return a.x*a.x+a.y*a.y;
    }
    friend bool operator<(const Point &a,const Point &b){
        if(fabs(a.y-b.y)<eps) return a.x<b.x;
        return a.y<b.y;
    }

};

double dis(Point a,Point b) {
    return sqrt((double)dis2(a-b));
}

int n,m;
char g[510][510];
bool vis[510][510];
int dir[8][2]= {{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
void dfs(int x,int y,int &lx,int &ly,int &rx,int &ry,Point &p1) {
    queue<Point>q;
    set<Point>s;
    s.insert(Point(x,y));
    q.push(Point(x,y));
    vis[x][y]=1;
    lx=x,ry=x,ly=y,ry=y;
    while(!q.empty()) {
        Point tem=q.front();
        q.pop();
        for(int i=0; i<8; i++) {
            int tx=tem.x+dir[i][0];
            int ty=tem.y+dir[i][1];
            if(tx>=1&&tx<=n&&ty>=1&&ty<=m&&!s.count(Point(tx,ty))&&g[tx][ty]!='.') {
                q.push(Point(tx,ty));
                s.insert(Point(tx,ty));
                vis[tx][ty]=1;
                lx=min(lx,tx);
                rx=max(rx,tx);
                ly=min(ly,ty);
                ry=max(ry,ty);
            }
        }
        p1=tem;
    }
}

int main() {
    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; i++) {
        scanf("%s",g[i]+1);
    }
    int M=0,S=0;
    for(int i=1; i<=n; i++) {
        for(int j=1; j<=m; j++) {
            Point p1,p2,p3,p4;
            int lx,ly,rx,ry;
            if(vis[i][j]) continue;
            if(g[i][j]=='.') continue;
            lx=i,rx=i,ly=j,ry=j;
            dfs(i,j,lx,ly,rx,ry,p1);
            p3=Point((rx+lx)/2,(ry+ly)/2);
            dfs(p1.x,p1.y,lx,ly,rx,ry,p2);
            p4=Point((p1.x+p2.x)/2,(p1.y+p2.y)/2);
            double len1=dis(p3,p4);
            double len2=dis(p1,p2);
            if(len1*2/len2<=0.5) S++;
            else M++;
        }
    }
    cout<<M<<" "<<S<<endl;
    return 0;
}