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;
}