[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;
}