• [TJOI2014]拼图


    嘟嘟嘟


    一眼看上去像状压dp,然后越想复杂度越不对劲,最后发现和爆搜差不多,索性就写爆搜了,复杂度(O()能过())
    别忘了填拼图和回溯的时候只动拼图中是1的部分,不要把(n * m)的矩形全改了。

    #include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<algorithm>
    #include<cstring>
    #include<cstdlib>
    #include<cctype>
    #include<vector>
    #include<stack>
    #include<queue>
    using namespace std;
    #define enter puts("") 
    #define space putchar(' ')
    #define Mem(a, x) memset(a, x, sizeof(a))
    #define In inline
    typedef long long ll;
    typedef double db;
    const int INF = 0x3f3f3f3f;
    const db eps = 1e-8;
    const int maxn = 17;
    const int N = 4;
    inline ll read()
    {
      ll ans = 0;
      char ch = getchar(), last = ' ';
      while(!isdigit(ch)) last = ch, ch = getchar();
      while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
      if(last == '-') ans = -ans;
      return ans;
    }
    inline void write(ll x)
    {
      if(x < 0) x = -x, putchar('-');
      if(x >= 10) write(x / 10);
      putchar(x % 10 + '0');
    }
    
    int n;
    struct Puzzle
    {
      int n, m, num;
      char a[maxn][maxn];
    }t[maxn];
    
    In void calc(int x)
    {
      for(int i = 1; i <= t[x].n; ++i)
        for(int j = 1; j <= t[x].m; ++j)
          if(t[x].a[i][j] == '1') ++t[x].num;
    }
    
    int f[maxn][maxn], Ans[maxn][maxn], Flg = -1;
    In void copy_ans()
    {
      for(int i = 1; i <= N; ++i)
        for(int j = 1; j <= N; ++j) Ans[i][j] = f[i][j];
    }
    In bool check(int now, int x, int y)
    {
      for(int i = x; i <= x + t[now].n - 1; ++i)
        for(int j = y; j <= y + t[now].m - 1; ++j)
          if(f[i][j] && t[now].a[i - x + 1][j - y + 1] == '1') return 0;
      return 1;
    }
    In void copy(int now, int x, int y, bool flg)
    {
      for(int i = 1; i <= t[now].n; ++i)
        for(int j = 1; j <= t[now].m; ++j)
          if(t[now].a[i][j] == '1')
    	f[i + x - 1][j + y - 1] = flg ? now : 0;
    }
    In void dfs(int now)
    {
      if(Flg > 0) return;
      if(now == n + 1) {++Flg; copy_ans(); return;}
      for(int i = 1; i <= N - t[now].n + 1; ++i)
        for(int j = 1; j <= N - t[now].m + 1; ++j)
          if(check(now, i, j))
    	{
    	  copy(now, i, j, 1);
    	  dfs(now + 1);
    	  copy(now, i, j, 0);
    	}
    }
    
    int main()
    {
      while(scanf("%d", &n) != EOF)
        {
          Flg = -1; Mem(f, 0);
          int sum = 0;
          for(int i = 1; i <= n; ++i)
    	{
    	  t[i].n = read(), t[i].m = read(); t[i].num = 0;
    	  for(int j = 1; j <= t[i].n; ++j) scanf("%s", t[i].a[j] + 1);
    	  calc(i); sum += t[i].num;
    	}
          if(sum ^ 16) {puts("No solution"); continue;}
          dfs(1);
          if(Flg == -1) puts("No solution");
          else if(Flg > 0) puts("Yes, many!");
          else
    	{
    	  puts("Yes, only one!");
    	  for(int i = 1; i <= N; ++i)
    	    {
    	      for(int j = 1; j <= N; ++j) write(Ans[i][j]);
    	      enter;
    	    }
    	}
        }
      return 0;
    }
    
  • 相关阅读:
    IEEE 754 浮点数的表示方法
    .NET Core 3.0及以上的EFCore连接MySql
    一些常见错误/技巧/结论总结
    2-sat学习笔记
    动态DP学习笔记
    动态规划优化算法——wqs二分 and 折线优化
    扩展莫队小总结(二) (回滚莫队/二次离线莫队)
    CF1504X Codeforces Round #712
    CF1500D Tiles for Bathroom (递推+大讨论)
    CF1486X Codeforces Round #703
  • 原文地址:https://www.cnblogs.com/mrclr/p/10466085.html
Copyright © 2020-2023  润新知