[USACO5.1]夜空繁星Starry Night

这个世界上怎么会有这么多恶心人的搜索啊

做法...

找联通块，找完之后尝试赋颜色

判重很玄学，算联通块上点的两两距离，求和。如果和一样，那么图形一定一样

然后就苦逼搜索就行了

1.0h

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath> 
#include<iostream>
#include<map>
using namespace std;
#define O(x) cout << #x << " " << x << endl;
#define O_(x) cout << #x << " " << x << "  ";
#define B cout << "breakpoint" << endl;
#define clr(a) memset(a,0,sizeof(a));
#define pii pair<int,int>
#define mp make_pair
typedef double db;
typedef long long ll;
inline int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') op = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        (ans *= 10) += ch - '0';
        ch  = getchar();
    }
    return ans * op;
}
const int maxn = 505;
const db eps = 0.00000001;
int w,h;
int g[maxn][maxn];
int mark[maxn][maxn];
int dx[maxn][maxn],dy[maxn][maxn],n2,n1;
db dist[maxn];
int ch = 0;
int c[maxn];
db calc(int x1,int y1,int x2,int y2)
{
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
int judge(int k)
{
  for(int i = 1;i <= n2;i++)
    for(int j = 1;j <= n2;j++)
    dist[k] += calc(dx[n1][i],dy[n1][i],dx[n1][j],dy[n1][j]);
  for(int i = 1;i < n1;i++) if(fabs(dist[i] - dist[k]) <= eps) return i;
  return 0;
}
void dfs(int x,int y)
{
  for(int i = -1;i <= 1;i++)
    for(int j = -1;j <= 1;j++)
      {
    int nx = x + i,ny = y + j;
    if(nx < 1 || nx > w || ny < 1 || ny > h || mark[nx][ny] || !g[nx][ny]) continue;
    mark[nx][ny] = ch;
    dx[n1][++n2] = nx,dy[n1][n2] = ny;
    dfs(nx,ny);
      }
}
int main()
{
  h = read(),w = read();
  for(int i = 1;i <= w;i++)
    for(int j = 1;j <= h;j++)
    {
        char cha = getchar();
        while(cha != '0' && cha != '1') cha = getchar();
        g[i][j] = cha - '0';
    }
  for(int i = 1;i <= w;i++)
    for(int j = 1;j <= h;j++)
      {
    if(!mark[i][j] && g[i][j])
      {
          ch++;
        n2 = 1,n1++;
        mark[i][j] = ch;
        dx[n1][n2] = i,dy[n1][n2] = j;
        c[n1] = ch;
        dfs(i,j);
        int k = judge(n1);
        if(k)
          {
        ch--;
        for(int i = 1;i <= n2;i++) mark[dx[n1][i]][dy[n1][i]] = c[k];
          }
      }
      }
    for(int i = 1;i <= w;i++)
    {
        for(int j = 1;j <= h;j++)
            mark[i][j] > 0 ? printf("%c",mark[i][j] + 'a' - 1) : printf("0");
        printf("
");
    }
}