http://poj.org/problem?id=1077
Source Code
Problem: 1077 User: p_hoenix
Memory: 220K Time: 0MS
Language: C++ Result: Accepted
Source Code
#include <iostream>
#include <utility>
#include <vector>
#include <fstream>
#include <math.h>
using namespace std;
const int MAX = 1 << 20;
const int ROW = 3, COL = 3;
//棋盘
struct Board
{
Board& operator=(const Board &b)
{
for(int i = 0; i < ROW; i++)
for(int j = 0; j < COL; j++)
tile[i][j] = b.tile[i][j];
r = b.r;
c = b.c;
return *this;
}
int tile[ROW][COL];//4*4的方块
int r,c;//空格的位置
};
typedef Board Status;
struct Step
{
int r,c;
int dir;
};
class PicturePizzle
{
public:
friend void input(PicturePizzle &pp);
PicturePizzle()
{
pStepDir = (int*)malloc(sizeof(int) * MAX);
stepHeap = (Board*)malloc(sizeof(Board) * MAX);
}
~PicturePizzle()
{
free(pStepDir);
free(stepHeap);
}
//移动步骤.
typedef struct{
int r,c;//空格位置
int dir;//移动方向
}Step;
//判断以该状态开始是否可以完成拼图
bool JudgeValidStatus();
//heuristic function,启发式函数,计算(估算)最大移动步数
int Heuristic();
//IDA*搜索寻找移动方案
//当前状态,深度,h(),前一步的移动方向.
bool IDAstar(Status &s, int depth, int h, int prevDir);
//打印移动路径
void ShowStep(int depth);
void Run();
//private:
Board board;
int h;
int maxDepth;
vector<Board> step;
int *pStepDir;
Board *stepHeap;
//郁闷的地方,C++ primer里面说可以给const static在里面初始化,在类外面定义的,但是VC6.0不可以
//目标棋盘
const static Board destBoard;// = {1 ,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 4, 4};
const static int nDestPos[ROW * COL + 1][2];//最终棋盘的各个方块的位置
const static int dir[4][2];
};
//1到16的方块编号
const Board PicturePizzle::destBoard = {1 ,2, 3, 4, 5, 6, 7, 8, 9, 2, 2};//10, 11, 12, 13, 14, 15, 16, 3, 3};
const int PicturePizzle::nDestPos[ROW * COL + 1][2] = {{-1, -1},
{0, 0}, {0, 1}, {0, 2},
{1, 0}, {1, 1},{1, 2},
{2, 0}, {2, 1}, {2, 2}};
//{3, 0}, {3, 1}, {3, 2}, {3, 3}};
const int PicturePizzle::dir[4][2] = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};//左上右下
bool PicturePizzle::JudgeValidStatus()
{
int i, j, array[ROW * COL], k, sum;
sum = k = 0;
//转化成一维数组
for(i = 0; i < ROW; i++)
for(j = 0; j < COL; j++)
{
array[k++] = board.tile[i][j];
}
//因为在最终状态下,棋盘的一维数组是有序递增的.
//除了X结点外,要到达目标位置需要移动的最大步数,因为在转换成了一维后原本可以上下移动的都被转化成了水平移动.
//但是幸运的是,他们是同奇同偶的.具体的证明我也不知道,画图有这么个规律
for(i = 0; i <= ROW * COL - 2; i++)
for(j = i + 1; j <= ROW * COL - 1; j++)
if(array[i] > array[j])
sum++;
//这个是X方块到目标位置的最短步数,不管怎么移动,只要最后是在目标位置,必定是同奇同偶的.(简单的考虑就是撤销)
//而每一次的其他方块的移动都是与X方块的交换来实现的,所以他们的和必定是偶数
sum += abs(board.r - nDestPos[ROW * COL][0]) + abs(board.c - nDestPos[ROW * COL][1]);
return sum % 2 == 0;
}
int PicturePizzle::Heuristic()
{
int r, c, tile, sum = 0;
for(r = 0; r < ROW; r++)
for(c = 0; c < COL; c++)
{
tile = board.tile[r][c];
//计算每个方块(非空格)移动到目标位置的曼哈顿距离,不需要计算空格
if(tile != ROW * COL)
{
sum += abs(r - nDestPos[tile][0]) + abs(c - nDestPos[tile][1]);
}
}
h = sum;
return sum;
}
int global = 0;
bool PicturePizzle::IDAstar(Status &s, int depth, int h, int prevDir)
{
//cout<<global++<<endl;
if(memcmp(&s, &destBoard, sizeof(destBoard)) == 0)//到达目标状态
{
this->ShowStep(depth);
return true;
}
if(depth >= maxDepth)//超过当前深度
return false;
Status ns;
int nr,nc,nh,moveTile;
for(int d = 0; d < 4; d++)
{
if(abs(d - prevDir) == 2)//会撤销上次的移动
continue;
ns = s;
nr = s.r + dir[d][0];
nc = s.c + dir[d][1];
if(nr < 0 || nr > ROW - 1 || nc < 0 || nc > COL - 1)//越界
continue;
//移动方块
moveTile = s.tile[nr][nc];
ns.tile[s.r][s.c] = moveTile;
ns.tile[nr][nc] = ROW * COL;
ns.r = nr;
ns.c = nc;
//向左移动.并且与X交换的方块离它的目标位置近了一步.
//反过来,因为X是向左的,那么moveTile就是向右的,那么这个时候只有moveTile在它的目标位置的左边的时候才成立
//也就是说它原来的位置nc是在目标的左边
if(d == 0 && nc < nDestPos[moveTile][1])
nh = h - 1;//nh比当前的h少1
else if(d == 1 && nr < nDestPos[moveTile][0])//上
nh = h - 1;
else if(d == 2 && nc > nDestPos[moveTile][1])//右
nh = h - 1;
else if(d == 3 && nr > nDestPos[moveTile][0])
nh = h - 1;
else
nh = h + 1;
//大于f
if(depth + nh > maxDepth)
continue;
//step.insert(step.begin() + depth, ns);
//step.erase(step.begin() + depth + 1);
pStepDir[depth] = d;
stepHeap[depth] = ns;
if(IDAstar(ns, depth + 1, nh, d))
return true;
}
return false;
}
void PicturePizzle::ShowStep(int depth)
{
int i = 0;
//cout<<depth<<endl;
for(i = 0; i < depth; i++)
switch(pStepDir[i])
{
case 0:
cout<<'l';
break;
case 1:
cout<<'u';
break;
case 2:
cout<<'r';
break;
case 3:
cout<<'d';
break;
}
cout<<endl;
/*
for(i = 0; i < depth; i++)
{
for(int r = 0; r < ROW; r++)
{
for(int c = 0; c < COL; c++)
cout<<stepHeap[i].tile[r][c]<<"\t";
cout<<endl;
}
cout<<endl<<endl;
}
*/
/*
cout<<step.size()<<endl<<endl;
for(vector<Board>::iterator iter = step.begin(); iter != step.end(); iter++)
{
cout<<i++<<endl;
for(int r = 0; r < 4; r++)
{
for(int c = 0; c < 4; c++)
cout<<iter->tile[r][c]<<"\t";
cout<<endl;
}
cout<<endl<<endl;
}*/
}
void PicturePizzle::Run()
{
if(JudgeValidStatus())
{
maxDepth = Heuristic();
while(maxDepth++)
if(IDAstar(board, 0, h, 20))
break;
}
else
{
cout<<"unsolvable"<<endl;
}
}
void input(PicturePizzle &pp)
{
//cout<<"输入一个文件(.txt),表示拼图的初始"
//fstream file("in.txt");
int num;
for(int r = 0; r < ROW; r++)
for(int c = 0; c < COL; c++)
{
//file>>num;
cin>>num;
if(/*file.fail()*/cin.fail())
{
//file.clear();
//file.ignore(1,'\t');
cin.clear();
cin.ignore(1, ' ');
pp.board.tile[r][c] = ROW * COL;
pp.board.r = r;
pp.board.c = c;
}
else
pp.board.tile[r][c] = num;
}
//file.close();
}
int main()
{
//freopen("out.txt", "w", stdout);
PicturePizzle pp;
input(pp);
pp.Run();
return 0;
}
