HDU 3088 WORM [BFS]

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Description

一个虫子，有三种颜色，rgb，它可以执行一种变换，两个不同的颜色变成一个相同的颜色，问，能不能变成全是一种颜色

Algorithm

BFS，每次变变变
同时用三进制r = 0, g = 1, b = 2来记录每一次的情况。

Code

#include <cstring>#include <iostream>#include <queue>using namespace std;struct V{  string s;  int step, h;};const int maxh = 1e6;bool h[maxh];int main(){  int n;  cin >> n;  for (int i = 0; i < n; i++)  {    memset(h, 0, sizeof(h));    string s;    cin >> s;    queue<V> q;    V t;    t.s = s;    t.step = 0;    int ss = 1; // 阶    int hh = 0; // 3进制表示值    for (int j = 0; j < s.size(); j++)    {      if (s[j] == 'g') hh += ss;      if (s[j] == 'b') hh += ss * 2;      ss *= 3;    }    h[hh] = true;    q.push(t);    int ans = -1;    while (!q.empty())    {      V now = q.front();      q.pop();      //检验是否同色      bool flag = true;      for (int j = 1; j < now.s.size(); j++)        if (now.s[j] != now.s[j - 1])        {          flag = false;          break;        }      if (flag)      {        ans = now.step;        break;      }      //变变变      for (int j = 1; j < now.s.size(); j++)        if (now.s[j] != now.s[j - 1])        {          V t;          t.s = now.s;          char ch;          if (now.s[j] == 'r' && now.s[j - 1] == 'g') ch = 'b';          if (now.s[j] == 'g' && now.s[j - 1] == 'r') ch = 'b';          if (now.s[j] == 'r' && now.s[j - 1] == 'b') ch = 'g';          if (now.s[j] == 'b' && now.s[j - 1] == 'r') ch = 'g';          if (now.s[j] == 'g' && now.s[j - 1] == 'b') ch = 'r';          if (now.s[j] == 'b' && now.s[j - 1] == 'g') ch = 'r';          t.s[j] = ch;          t.s[j - 1] = ch;          ss = 1; // 阶          hh = 0; // 3进制表示          for (int k = 0; k < t.s.size(); k++)          {            if (t.s[k] == 'g') hh += ss;            if (t.s[k] == 'b') hh += ss * 2;            ss *= 3;          }          if (!h[hh])          {            h[hh] = true;            t.step = now.step + 1;            q.push(t);          }        }    }    if (ans == -1) cout << "No solution!" << endl; else cout << ans << endl;  }}

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