The Rotation Game UVA - 1343（IDA* 状态空间搜索）

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int line[8][7] =   // lines E~H are computed with the help of rev[] //用一系列“指针”将一维数组构造为特殊形态数据结构储存数据//修改时通过存储的特殊形态“指针”查询一维数组修改其内数据
{
{ 0, 2, 6,11,15,20,22 }, // A
{ 1, 3, 8,12,17,21,23 }, // B
{ 10, 9, 8, 7, 6, 5, 4 }, // C
{ 19,18,17,16,15,14,13 }, // D
};
const int rev[8] = { 5, 4, 7, 6, 1, 0, 3, 2 }; // reverse lines of each line

const int center[8] = { 6, 7, 8, 11, 12, 15, 16, 17 };

int a[25]; //状态

bool final()
{
for (int i = 0; i < 8; i++)
if (a[center[i]] != a[center[0]])
return false;
return true;
}

char ans[1000];

inline int diff(int n)
{
int ans = 0;
for (int i = 0; i < 8; i++)
{
if (a[center[i]] != n)
ans++;
}
return ans;
}

inline int h()
{
return min(min(diff(1), diff(2)), diff(3)); //中间8个格子中最少的相同格子数
}

inline void move(int dir)
{
int tmp = a[line[dir][0]];
for (int i = 0; i < 6; i++)
{
a[line[dir][i]] = a[line[dir][i + 1]];
}
a[line[dir][6]] = tmp;
}

bool dfs(int d, int maxd)
{
if (final())
{
ans[d] = '\0'; //以字符串格式符输出，需在字符串数组尾部添加'\0'
printf("%s\n", ans);
return true;
}
if (d + h() > maxd)
return false;
for (int i = 0; i < 8; i++)
{
ans[d] = 'A' + i; //打印路径，用数组存储解
move(i);
if(dfs(d+1, maxd))
return true;
move(rev[i]); //借助rev数组实现回溯操作
}
return false;
}

int main()
{
for (int i = 4; i < 8; i++)
{
for (int j = 0; j < 7; j++)
{
line[i][j] = line[rev[i]][6 - j]; //补全line数组
}
}
while (scanf("%d", &a[0]) && a[0])
{
for (int i = 1; i < 24; i++)
scanf("%d", &a[i]);
if (final())
{
printf("No moves needed\n");
}
else
{
for (int maxd = 1; ; maxd++) //IDA*不知最大深度，但一定有解时，不设max_ans
if (dfs(0, maxd))
break;
}
printf("%d\n", a[6]);//
}
}