世界名画陈列馆问题

#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;

const int MAX = 50;
int board[MAX][MAX];  //记录方格被监视情况
int root[MAX][MAX];      //记录机器人位置
int best[MAX][MAX];
int m, n;             //矩阵为 m * n
int k = 0;                //机器人个数
int bestk = 10000;
int t = 0;            //被监视的方格个数
int t1, t2, more;
int dx[] = {0, 1, -1, 0, 0}; //原位置及四个方向
int dy[] = {0, 0, 0,  1, -1};

void change(int i, int j)
{
root[i][j] = 1;  //放置机器人
k++;
for(int d=0; d<=4; d++)     //原位置及周围被监视
{
int p = i + dx[d];
int q = j + dy[d];
board[p][q]++;
if(board[p][q] == 1)
t++;
}
}

//恢复被监视前的状态
void restore(int i, int j)
{
root[i][j] = 0;
k--;
for(int d=0; d<=4; d++)
{
int p = i + dx[d];
int q = j + dy[d];
board[p][q]--;
if(board[p][q] == 0)
t++;
}
}

void backtrack(int i, int j)
{
do  //找到未被监视的方格
{
j++;
if(j > n)
{
i++;
j = 1;
}
}while(board[i][j] != 0 && i<=n);

if(i>n)  //到达叶结点，复制最优解
{
if(k < bestk)
{
bestk = k;
for(int i=1; i<=m; i++)
for(int j=1; j<=n; j++)
best[i][j] = root[i][j];
}
}
if(k + (t1 - t)/5 >= bestk)  //剪枝
return;
if(i< n-1 && (k + (t2 - t) / 5 >= bestk))  //剪枝
return;
if(i<n)
{
change(i+1, j);  //放置机器人，并改变状态
backtrack(i, j);
restore(i+1, j); //撤销机器人，并恢复状态
}
if(j<m && (board[i][j+1]==0 || board[i][j+2]==0))
{
change(i, j+1);
backtrack(i, j);
restore(i, j+1);
}
if(board[i+1][j]==0 && board[i][j+1]==0)
{
change(i, j);
backtrack(i, j);
restore(i, j);
}
}

void compute()
{
more = m / 4 + 1;
if(m % 4 == 3)
more++;
else if(m % 4 == 2)
more += 2;
t2 = m * n + more + 4;
t1 = m * n + 4;
memset(board, 0, sizeof(board));
memset(root, 0, sizeof(root));
for(int i=0; i<=n+1; i++)      //最外层一圈为1
{
board[i][m+1] = 1;
board[i][0] = 1;
}
for(i=0; i<=m+1; i++)
{
board[n+1][i] = 1;
board[n+1][0] = 1;
}

if(n==1 && m==1)
cout << 1 << endl;
else
backtrack(1, 0);
}

int main()
{
ifstream fin("名画陈列馆.txt");
cout << "输入矩阵行数：";
fin >> m;   cout << m << endl;
cout << "输入矩阵列数：";
fin >> n;   cout << n << endl;

compute();

cout << "最少的机器人个数为：" << bestk << endl;
cout << "机器人位置为：\n";
for(int i=1; i<=m; i++)
{
for(int j=1; j<=n; j++)
cout << best[i][j] << "\t";
cout << endl;
}
cout << endl;
fin.close();
return 0;
}