喷漆机器人问题

#include <iostream>#include <fstream>#include <cmath>using namespace std;const int MAX = 200;const int INF = 100000;int color[MAX];   //存储每个小矩形的颜色int power2[MAX];  //存储与小矩形对应的每一位bool under[MAX][MAX];  //under[i][j]=true 表示小矩形j在小矩形i的下方int num[MAX][INF];  //num[i][np]表示当前对小矩形i涂漆，剩余的小矩形对应值np中相应的位未涂漆，机器人拿起喷枪的次数int board[MAX][MAX];  //存储小矩形编号int n;void backtrack(int dep, int p){    if(num[dep][p] >= 0)        return;    int i;    for(i=0; i<n; i++)    {        if(dep!=i && (p & power2[i])>0 && under[i][dep]) //如果i,dep不是同一个矩形，i仍未涂漆，i在dep的上方        {            num[dep][p] = INF;  //失败            return;        }    }    int np = p - power2[dep];  //化为二进制，1代表相应的小矩形未涂漆    if(np == 0)  //没有未涂漆的小矩形        num[dep][p] = 1;    else        for(i=0; i<n; i++)  //寻找下一个未涂漆的小矩形        {            if((np & power2[i]) > 0)            {                 backtrack(i, np);                int v = num[i][np] + (color[dep]==color[i]? 0:1);                 if(num[dep][p]<0 || num[dep][p]>v)                    num[dep][p] = v;             }        }}//执行回溯法，输出最优值void search(){    int i, j;    memset(num, -1, sizeof(num));    int opt = -1;  //拿起喷枪最少次数    for(i=0; i<n; i++)  //尝试将每一个矩形作为最开始喷漆的矩形    {        backtrack(i, power2[n]-1);        if(opt<0 || opt>num[i][power2[n]-1])            opt = num[i][power2[n]-1];    }    cout << "机器人拿起喷枪最少次数为：" << opt;}int main(){    ifstream fin("喷漆机器人.txt");    cout << "输入小矩形个数：";    fin >> n;   cout << n << endl;    cout << "输入各个小矩形的左上角、右下角坐标、颜色\n";    int x1, y1, x2, y2, i, j, x, y;    memset(board, -1, sizeof(board));    for(i=0; i<n; i++)    {        fin >> x1 >> y1 >> x2 >> y2 >> color[i];        cout << x1 << " " << y1 << " " << x2 << " " << y2 << " " << color[i] << endl;        for(x=x1; x<x2; ++x)            for(y=y1; y<y2; ++y)                board[x][y] = i;  //为小矩形编号          }    power2[0] = 1;    for(i=1; i<=n; i++) //存储每一个小矩形对应的位        power2[i] = power2[i-1] << 1;    memset(under, false, sizeof(under));    for(i=0; i<MAX; i++)        for(j=0; j<MAX; j++)            if(board[i][j]>=0 && board[i+1][j]>=0 && board[i+1][j]!=board[i][j])                under[ board[i][j] ][ board[i+1][j] ] = true;  // 矩形board[i+1][j]在矩形board[i][j]下方    search();    cout << endl;    cout << endl;    fin.close();    return 0;}