1 //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
2 //看了讨论区写了一发
3 //套一下思路：
4 //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。
5 #include <bits/stdc++.h>
6 using namespace std;
7 #define LL long long
8 typedef pair<int,int> pii;
9 const int inf = 0x3f3f3f3f;
10 const int N =1e5+10;
11 #define clc(a,b) memset(a,b,sizeof(a))
12 const double eps = 1e-8;
13 const int MOD = 1e9+7;
14 void fre() {freopen("in.txt","r",stdin);}
15 void freout() {freopen("out.txt","w",stdout);}
16 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;}
17
18 int sgn(double x) {
19 if(fabs(x) < eps)return 0;
20 if(x < 0)return -1;
21 else return 1;
22 }
23
24 struct Point {
25 int x,y;
26 Point() {}
27 Point(int _x,int _y) {
28 x = _x;
29 y = _y;
30 }
31 Point operator -(const Point &b)const {
32 return Point(x - b.x,y - b.y);
33 }
34 int operator ^(const Point &b)const {
35 return x*b.y - y*b.x;
36 }
37 int operator *(const Point &b)const {
38 return x*b.x + y*b.y;
39 }
40 friend int dis2(Point a) {
41 return a.x*a.x+a.y*a.y;
42 }
43 friend bool operator<(const Point &a,const Point &b){
44 if(fabs(a.y-b.y)<eps) return a.x<b.x;
45 return a.y<b.y;
46 }
47
48 };
49
50 double dis(Point a,Point b) {
51 return sqrt((double)dis2(a-b));
52 }
53
54 int n,m;
55 char g[510][510];
56 bool vis[510][510];
57 int dir[8][2]= {{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
58 void dfs(int x,int y,int &lx,int &ly,int &rx,int &ry,Point &p1) {
59 queue<Point>q;
60 set<Point>s;
61 s.insert(Point(x,y));
62 q.push(Point(x,y));
63 vis[x][y]=1;
64 lx=x,ry=x,ly=y,ry=y;
65 while(!q.empty()) {
66 Point tem=q.front();
67 q.pop();
68 for(int i=0; i<8; i++) {
69 int tx=tem.x+dir[i][0];
70 int ty=tem.y+dir[i][1];
71 if(tx>=1&&tx<=n&&ty>=1&&ty<=m&&!s.count(Point(tx,ty))&&g[tx][ty]!='.') {
72 q.push(Point(tx,ty));
73 s.insert(Point(tx,ty));
74 vis[tx][ty]=1;
75 lx=min(lx,tx);
76 rx=max(rx,tx);
77 ly=min(ly,ty);
78 ry=max(ry,ty);
79 }
80 }
81 p1=tem;
82 }
83 }
84
85 int main() {
86 scanf("%d%d",&n,&m);
87 for(int i=1; i<=n; i++) {
88 scanf("%s",g[i]+1);
89 }
90 int M=0,S=0;
91 for(int i=1; i<=n; i++) {
92 for(int j=1; j<=m; j++) {
93 Point p1,p2,p3,p4;
94 int lx,ly,rx,ry;
95 if(vis[i][j]) continue;
96 if(g[i][j]=='.') continue;
97 lx=i,rx=i,ly=j,ry=j;
98 dfs(i,j,lx,ly,rx,ry,p1);
99 p3=Point((rx+lx)/2,(ry+ly)/2);
100 dfs(p1.x,p1.y,lx,ly,rx,ry,p2);
101 p4=Point((p1.x+p2.x)/2,(p1.y+p2.y)/2);
102 double len1=dis(p3,p4);
103 double len2=dis(p1,p2);
104 if(len1*2/len2<=0.5) S++;
105 else M++;
106 }
107 }
108 cout<<M<<" "<<S<<endl;
109 return 0;
110 }
