P6201 & P1985 Fliptile S 题解

合集 - 题解(48)

1.CSES 1667 Message Route 题解2023-02-132.P3956 棋盘 题解2023-02-223.abc233_f Swap and Sort 题解2023-03-034.abc233_e Σ[k=0..10^100]floor(X／10^k) 题解2023-03-035.abc233_d Interval 题解2023-03-036.abc233_c Product 题解2023-03-017.abc249_f Ignore Operations 题解2023-04-168.abc249_d Index Trio 题解2023-04-169.abc248_e K-colinear Line 题解2023-04-1310.abc247_f Cards 题解2023-04-1211.CF1066C Books Queries 题解2023-03-1412.P1038 神经网络 题解2023-03-1413.SeekLuna P1362 拓扑排序 3 题解2023-03-1414.abc232_e Rook Path 题解2023-03-0315.abc235_e MST + 1 题解2023-05-1716.abc234_e Arithmetic Number 题解2023-05-1717.P8786 李白打酒加强版 题解2023-05-1718.abc235_d Multiply and Rotate 题解2023-05-1719.CF1095D Circular Dance 题解2023-05-17

20.P6201 & P1985 Fliptile S 题解2023-05-17

21.CF1183C Computer Game 题解2023-05-1722.CF1095E Almost Regular Bracket Sequence 题解2023-05-1723.abc256_e Takahashi's Anguish 题解2023-05-1724.abc260_g Scalene Triangle Area 题解2023-05-1625.P8714 填空问题 题解2023-05-0926.abc252_d Distinct Trio 题解2023-04-2927.abc252_f Bread 题解2023-04-2328.abc253_e Distance Sequence 题解2023-04-2329.abc250_e Prefix Equality 题解2023-04-1630.abc250_d 250-like Number 题解2023-04-1631.arc164_a Ternary Decomposition 题解2023-07-1232.abc275_f Erase Subarrays 题解2023-06-0233.abc275_e Sugoroku 4 题解2023-05-3134.abc274_d Robot Arms 2 题解2023-05-3135.abc260_f Find 4-cycle 题解2023-05-2436.abc260_e At Least One 题解2023-05-2437.abc273_e Notebook 题解2023-05-2438.abc271_f XOR on Grid Path 题解2023-05-2439.abc271_e Subsequence Path 题解2023-05-2240.abc271_c Manga 题解2023-05-2241.abc269_f Numbered Checker 题解2023-05-1842.abc270_f Transportation 题解2023-05-1743.CF1077E Thematic Contests 题解2023-05-1744.CF1935D Exam in MAC 题解2024-04-0145.CF1144G Two Merged Sequences 题解2024-06-1546.joi2022_yo2_c 国土分割 (Land Division) 题解2024-07-0947.P3588 PUS 题解2024-07-2048.CF1946F Nobody is needed 题解2024-08-03

收起

Fliptile S & 翻转棋 Fliptile S

暴力-47pts

搜索每个格子翻或不翻，最后判断是否每个点都为零即可

点击查看代码

#include <bits/stdc++.h>
 
using namespace std;
 
const int dx[5] = {-1, 0, 0, 1, 0}, dy[5] = {0, -1, 1, 0, 0};
 
int n, m, a[20][20], ans[400], b[20][20], c[400], ansum;
bool f;
 
bool check(){
  for (int i = 1; i <= n; i++){
    for (int j = 1; j <= m; j++){
      if (b[i][j]){
        return 0;
      }
    }
  }
  return 1;
}
 
void dfs(int x, int y, int t, int sum){
  if (sum > ansum){
    return ;
  }
  if (f && sum == ansum && c[t - 1] > ans[t - 1]){
    return ;
  }
  if (x == n && y == m + 1){
    if (check()){
      if (!f || sum < ansum){
        for (int i = 1; i < t; i++){
          ans[i] = c[i];
        }
        f = 1;
        return ;
      }
      for (int i = 1; i < t; i++){
        if (c[i] < ans[i]){
          for (int j = 1; j < t; j++){
            ans[i] = c[i];
          }
          break;
        } else if (c[i] > ans[i]){
          break;
        }
      }
    }
    return ;
  }
  if (y == m + 1){
    dfs(x + 1, 1, t, sum);
    return ;
  }
  for (int i = 0; i < 5; i++){
    b[x + dx[i]][y + dy[i]] = !b[x + dx[i]][y + dy[i]];
  }
  c[t] = 1;
  dfs(x, y + 1, t + 1, sum + 1);
  for (int i = 0; i < 5; i++){
    b[x + dx[i]][y + dy[i]] = !b[x + dx[i]][y + dy[i]];
  }
  c[t] = 0;
  dfs(x, y + 1, t + 1, sum);
}
 
int main() {
  cin >> n >> m;
  ansum = n * m + 1;
  for (int i = 1; i <= n; i++){
    for (int j = 1; j <= m; j++){
      cin >> a[i][j];
      b[i][j] = a[i][j];
    }
  }
  dfs(1, 1, 1, 0);
  if (!f){
    cout << "IMPOSSIBLE";
  } else {
    for (int i = 1; i <= n * m; i++){
      cout << ans[i] << ' ';
      if (i % m == 0){
        cout << '\n';
      }
    }
  }
  return 0;
}

正解-100pts

首先，在字典序最小的情况下，翻的次数只有两种

1. 翻奇数次相当于翻 $1$ 次
2. 翻偶数次相当于翻 $0$ 次

当第一排固定之后，后面翻或不翻都是固定的。

• 当第一排固定后，如果上一行的对应位置为 $1$，那么当前位置 必须 翻一次，依此类推

最后只需判断第 $M$ 行是否全为零即可。

点击查看代码

#include <bits/stdc++.h>
 
using namespace std;
 
const int dx[5] = {-1, 0, 0, 1, 0}, dy[5] = {0, -1, 1, 0, 0};
 
int n, m, a[20][20], ans[400], b[20][20], leb[20][20], c[400], sum, ansum;
bool f;
 
void s(int x, int sum){
  if (x == m + 1){
    int num = 0, t = m + 1;
    for (int i = 1; i <= n; i++){
      for (int j = 1; j <= m; j++){
        leb[i][j] = b[i][j];
      }
    }
    // 之后每行都是固定的
    for (int i = 2; i <= n; i++){
      for (int j = 1; j <= m; j++){
        if (b[i - 1][j]){
          num++, c[t] = 1;
          for (int k = 0; k < 5; k++){
            b[i + dx[k]][j + dy[k]] = !b[i + dx[k]][j + dy[k]];
          }
        } else {
          c[t] = 0;
        }
        t++;
      }
    }
    bool flag = 1;
    for (int i = 1; i <= m; i++){
      if (b[n][i]){ // 有内鬼!
        flag = 0;
        break;
      }
    }
    if (flag && sum + num < ansum){ // 没有内鬼并且翻的次数少些
      f = 1;
      ansum = sum + num;
      for (int i = 1; i <= n * m; i++){
        ans[i] = c[i];
      }
    }
    for (int i = 1; i <= n; i++){
      for (int j = 1; j <= m; j++){
        b[i][j] = leb[i][j];
      }
    }
    return ;
  }
  s(x + 1, sum); // 注意!先选不翻的情况才能使字典序最小
  // 翻!
  for (int j = 1; j < 5; j++){
    b[1 + dx[j]][x + dy[j]] = !b[1 + dx[j]][x + dy[j]];
  }
  c[x] = 1;
  s(x + 1, sum + 1);
  // 回溯
  c[x] = 0;
  for (int j = 1; j < 5; j++){
    b[1 + dx[j]][x + dy[j]] = !b[1 + dx[j]][x + dy[j]];
  }
}
 
int main() {
  cin >> n >> m;
  ansum = n * m + 1;
  for (int i = 1; i <= n; i++){
    for (int j = 1; j <= m; j++){
      cin >> a[i][j];
      b[i][j] = a[i][j];
    }
  }
  s(1, 0);
  if (!f){ // 不成立
    cout << "IMPOSSIBLE";
  } else {
    for (int i = 1; i <= n * m; i++){
      cout << ans[i] << ' ';
      if (i % m == 0){
        cout << '\n';
      }
    }
  }
  return 0;
}