[题解] luogu P1985 [USACO07OPEN]翻转棋

题面
今天学搜索,正好水一发以前做的这道毒瘤题

话说这道题做了我一天,别人都是各种优化,不超100行

天真的我硬核刚了220行,全程0优化水过

但其实不用这么长,我有的函数写的有点重复了(

思路:

显然是dfs,一行一行的来

搜到[i, j]时(i > 1),看[i - 1, j]是否为黑,是的话就翻转[i, j],

也就是说搜完当前行就要保证上一行的棋全都翻成了白色

当搜到最后一行时,

既要保证上一行翻成白色,还要保证自己也都翻成白色,

最后还要特判一下最后两个的翻转.

当时年少轻狂,我想着层次一定要清晰,
于是就SB地分别打了四个函数:

  1. 特判第一行,firstLineS
  2. 当只有一列的时候,dfs1
  3. 当搜到最后一行时或本来只有一行时,dfsn
  4. 正常搜索,dfs
这个题有个坑,92分卡了一下午: 搜到的第一个解不能直接输出,它是最优解,但不一定是字典序最小的,所以继续搜完

Code

#include <bits/stdc++.h>

using namespace std;

int n, m, a, minans = 99999;
int Chess[25][25], ans[25][25], minan[25][25];

void c(int i, int j) 
{
    if(Chess[i][j]%2 == 0) Chess[i][j] = 1;
    else Chess[i][j] = 0;
}
void print() 
{
    a = 0;
    for (int i = 1; i <= m; i++)
        for (int j = 1; j <= n; j++)
            if (Chess[i][j] % 2 == 1) return ;
    for (int i = 1; i <= m; i++)
        for (int j = 1; j <= n; j++)
            a += ans[i][j] % 2;
    if(minans > a) 
    {
        minans = a;
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                minan[i][j] = ans[i][j] % 2;
    }
}
int back1(int p) 
{
    if ( Chess[m][p] == 1 && p == n)
        return 1;
    else if (p != n)
        return 1;
    return 0;
}
int back2(int p) 
{
    if ( Chess[m][p] == 0 && p == n)
        return 0;
    else if (p != n)
        return 0;
    return 1;
}
void dfsn(int num) {    
    if (num == n + 1) {
        print();
        return ;
    }
    int line = m;
    if (m == 1)
    {
        if (num == 1)
        {
            dfsn(num + 1);    
            c(line, num + 1);
            c(line, num);
            ans[line][num]++;
            dfsn(num + 1);
        }
        if (Chess[line][num - 1]==0 && back2(num)==0)
            dfsn(num+1);
        else if (Chess[line][num - 1]==1 && back1(num)==1)
        {
            c(line, num - 1);
            c(line, num + 1);
            c(line, num);
            ans[line][num]++;
        
            dfsn(num + 1);
        }
    }
    else if (num == 1)
    {
        if (Chess[m - 1][1] == 0)
            dfsn(2);    
    
        else 
        {
            c(m - 1 , 1);
            c(m, 2);
            c(m, 1);
            ans[m][1]++;
        
            dfsn(2);
        }
    }
    
    else if (Chess[line - 1][num]==1 && Chess[line][num - 1]==1 && back1(num)==1)
    {
        c(line - 1 , num);
        c(line, num - 1);
        c(line, num + 1);
        c(line, num);
        ans[line][num]++;
        
        dfsn(num + 1);
        /*    
        ans[line][num]--;
        c(line - 1, num);
        c(line, num - 1);
        c(line, num + 1);
        c(line, num);
        */
    }
    else if (Chess[line - 1][num] == 0 && Chess[line - 1][num] == 0 && back2(num) == 0)
        dfsn(num + 1);
}
void dfs(int line, int num)
{
    if (num == n + 1 && line == m - 1)
    {
        dfsn(1);
        return ;
    }
    if (num == n + 1)
    {
        dfs(line + 1, 1);
        return ;
    }
    if (Chess[line - 1][num] == 0)
        dfs(line, num + 1);
    else
    {
        c(line - 1, num);
        c(line + 1, num);
        c(line, num - 1);
        c(line, num + 1);
        c(line, num);
        ans[line][num]++;
        dfs(line, num + 1);
 /*
        ans[line][num]--;        
        c(line - 1, num);
        c(line + 1, num);
        c(line, num - 1);
        c(line, num + 1);
        c(line, num);
*/
    }
}
void firstLineS(int k)
{
    if (k == n + 1) 
    {
        if (m > 2)
            dfs(2, 1);
        else if (m == 2)
            dfsn(1);
        return ;
    } 
    firstLineS(k + 1);
    c(1, k);
    c(1, k - 1);
    c(1, k + 1);
    c(2, k);
    ans[1][k]++;
    firstLineS(k + 1);
    ans[1][k]--;
    c(1, k);
    c(1, k - 1);
    c(1, k + 1);
    c(2, k);
}
void dfs1(int num)
{
    if (num == m + 1) 
    {
        print();
        return ;
    }
    if (num == 1)
    {
        dfs1(num + 1);
        ans[num][1]++;
        c(num, 1);
        c(num + 1, 1);
        dfs1(num + 1);
        ans[num][1]--;
        c(num, 1);
        c(num + 1, 1);
    }
    if(Chess[num - 1][1] == 0)
        dfs1(num + 1);
    else  {
        ans[num][1]++;
        c(num, 1);
        c(num + 1, 1);
        c(num - 1, 1);
        
        dfs1(num + 1); 
        
        ans[num][1]--;
        c(num + 1, 1);
        c(num - 1, 1);
        c(num, 1);
    }
}

int main()   
{
    cin >> m>> n;
    for (int i = 1; i <= m; i++)
        for (int j = 1; j <= n; j++)
            cin >> Chess[i][j];
    if (m == 1) dfsn(1);
    if (n == 1) dfs1(1);
    firstLineS(1);
    if (minans==99999)
        cout<<"IMPOSSIBLE"<<endl;
    else
        for (int i = 1; i <= m; i++)
        {    
            for (int j = 1; j <= n; j++)
                cout << minan[i][j]%2<<" ";
            cout << endl;
        }
    return 0;
}

Last modification:April 5th, 2019 at 09:48 am

4 comments

  1. Atlantis

    %%%

  2. fpjo

    TQL

  3. JRLC

    %%%%%%%%%%%%%%%%%%%%%%%

  4. Garbage

    抢前排

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