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  • luogu2346 四子连棋

    题目大意

      在一个4*4的棋盘上摆放了14颗棋子,其中有7颗白色棋子,7颗黑色棋子,有两个空白地带,任何一颗黑白棋子都可以向上下左右四个方向移动到相邻的空格,这叫行棋一步,黑白双方交替走棋,任意一方可以先走,如果某个时刻使得任意一种颜色的棋子形成四个一线(包括斜线),这样的状态为目标棋局。求用最少的步数移动到目标棋局的步数。

      总体思路很简单,Bfs即可,只是需要注意以下几点:

    • memcmp的返回值不一定是-1, 0, 1,而是<0, =0, >0的某个数。这在windows和linux上的效果不一样。
    • 注意:黑白双方交替走棋。
    • 任意一方都必须走一步。
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    #include <set>
    #include <cassert>
    using namespace std;
    
    const int MAX_N = 10;
    const int N = 4;
    const int Dir[4][2] = { {1, 0}, {0, 1}, {-1, 0}, {0, -1} };
    
    struct Node
    {
    	char A[MAX_N][MAX_N];
    	int Level;
    	char NextColor;
    
    	Node()
    	{
    		memset(A, 0, sizeof(A));
    		Level = 0;
    	}
    
    	Node operator = (const Node& a)
    	{
    		memcpy(A, a.A, sizeof(A));
    		Level = a.Level;
    		NextColor = a.NextColor;
    		return *this;
    	}
    
    	bool operator < (const Node& a) const
    	{
    		if (NextColor != a.NextColor)
    			return NextColor == 'B';
    		else
    			return memcmp(A, a.A, sizeof(A)) < 0;
    	}
    
    	bool operator == (const Node& a) const
    	{
    		return NextColor == a.NextColor && memcmp(A, a.A, sizeof(A)) == 0;
    	}
    
    	void OPos1(int &oRow1, int &oCol1)
    	{
    		for (int row = 1; row <= N; row++)
    			for (int col = 1; col <= N; col++)
    				if (A[row][col] == 'O')
    				{
    					oRow1 = row;
    					oCol1 = col;
    					return;
    				}
    	}
    
    	void OPos2(int &oRow2, int &oCol2)
    	{
    		int oRow1, oCol1;
    		OPos1(oRow1, oCol1);
    		for (int col = oCol1 + 1; col <= N; col++)
    			if (A[oRow1][col] == 'O')
    			{
    				oRow2 = oRow1;
    				oCol2 = col;
    				return;
    			}
    		for (int row = oRow1 + 1; row <= N; row++)
    			for (int col = 1; col <= N; col++)
    				if (A[row][col] == 'O')
    				{
    					oRow2 = row;
    					oCol2 = col;
    					return;
    				}
    		assert(0);
    	}
    
    	bool CanMove1(const int dRow, const int dCol)
    	{
    		int oRow1, oCol1;
    		OPos1(oRow1, oCol1);
    		int nextRow = oRow1 + dRow, nextCol = oCol1 + dCol;
    		return A[nextRow][nextCol] == NextColor && nextRow <= N && nextRow >= 1 && nextCol <= N && nextCol >= 1;
    	}
    
    	Node GetMove1(int dRow, int dCol)
    	{
    		int oRow1, oCol1;
    		OPos1(oRow1, oCol1);
    		Node ans = *this;
    		swap(ans.A[oRow1][oCol1], ans.A[oRow1 + dRow][oCol1 + dCol]);
    		return ans;
    	}
    
    	bool CanMove2(const int dRow, const int dCol)
    	{
    		int oRow2, oCol2;
    		OPos2(oRow2, oCol2);
    		int nextRow = oRow2 + dRow, nextCol = oCol2 + dCol;
    		return A[nextRow][nextCol] == NextColor && nextRow <= N && nextRow >= 1 && nextCol <= N && nextCol >= 1;
    	}
    
    
    	Node GetMove2(int dRow, int dCol)
    	{
    		int oRow2, oCol2;
    		OPos2(oRow2, oCol2);
    		Node ans = *this;
    		swap(ans.A[oRow2][oCol2], ans.A[oRow2 + dRow][oCol2 + dCol]);
    		return ans;
    	}
    
    	bool Ok()
    	{
    		for (int row = 1; row <= N; row++)
    		{
    			char st = A[row][1];
    			bool ok = true;
    			for (int col = 2; col <= N; col++)
    				if (A[row][col] != st)
    				{
    					ok = false;
    					break;
    				}
    			if (ok)
    				return true;
    		}
    		for (int col = 1; col <= N; col++)
    		{
    			char st = A[1][col];
    			bool ok = true;
    			for (int row = 2; row <= N; row++)
    				if (A[row][col] != st)
    				{
    					ok = false;
    					break;
    				}
    			if (ok)
    				return true;
    		}
    		char st = A[1][1];
    		bool ok = true;
    		for (int i = 2; i <= N; i++)
    			if (A[i][i] != st)
    			{
    				ok = false;
    				break;
    			}
    		if (ok)
    			return true;
    		st = A[1][N];
    		ok = true;
    		for (int row = 2, col = N - 1; row <= N; row++, col--)
    			if (A[row][col] != st)
    			{
    				ok = false;
    				break;
    			}
    		return ok;
    	}
    };
    Node Start;
    
    int Bfs()
    {
    	static queue<Node> q;
    	static set<Node> cache;
    	Node s1 = Start, s2 = Start;
    	s1.NextColor = 'B';
    	s2.NextColor = 'W';
    	q.push(s1);
    	q.push(s2);
    	cache.insert(s1);
    	cache.insert(s2);
    	while (!q.empty())
    	{
    		Node cur = q.front();
    		q.pop();
    		if (!(cur == s1 || cur == s2) && cur.Ok())
    			return cur.Level;
    		for (int i = 0; i < 4; i++)
    		{
    			if (cur.CanMove1(Dir[i][0], Dir[i][1]))
    			{
    				Node next = cur.GetMove1(Dir[i][0], Dir[i][1]);
    				next.NextColor = (cur.NextColor == 'B' ? 'W' : 'B');
    				if (!cache.count(next))
    				{
    					next.Level = cur.Level + 1;
    					cache.insert(next);
    					q.push(next);
    				}
    			}
    			if (cur.CanMove2(Dir[i][0], Dir[i][1]))
    			{
    				Node next = cur.GetMove2(Dir[i][0], Dir[i][1]);
    				next.NextColor = (cur.NextColor == 'B' ? 'W' : 'B');
    				if (!cache.count(next))
    				{
    					next.Level = cur.Level + 1;
    					cache.insert(next);
    					q.push(next);
    				}
    			}
    		}
    	}
    	return -1;
    }
    
    int main()
    {
    	for (int i = 1; i <= 4; i++)
    		scanf("%s", Start.A[i] + 1);
    	printf("%d
    ", Bfs());
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/headboy2002/p/9665652.html
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