IDA*

IDA*

IDA* + 迭代加深
当前深度 + 估价函数 > 深度限制 ---> return；

180. 排书


样例输入：

3
6
1 3 4 6 2 5
5
5 4 3 2 1
10
6 8 5 3 4 7 2 9 1 10

样例输出：

2
3
5 or more

代码模板：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 15;

int n;
int q[N], w[5][N];

int f() //估价函数
{
    int res = 0;
    for (int i = 0; i + 1 < n; i ++ )
        if (q[i + 1] != q[i] + 1)
            res ++ ;
    return (res + 2) / 3;
}

bool check()  
{
    for (int i = 0; i < n; i ++ )
        if (q[i] != i + 1)
            return false;
    return true;
}
 //depth当前深度   max_depth最大深度
bool dfs(int depth, int max_depth)
{
    if (depth + f() > max_depth) return false; //当前深度加上估价函数  
    if (check()) return true;  //是否升序

    for (int l = 0; l < n; l ++ )
        for (int r = l; r < n; r ++ )
            for (int k = r + 1; k < n; k ++ )
            {
                //备份
                memcpy(w[depth], q, sizeof q);
                int x, y;
                //位置交换
                for (x = r + 1, y = l; x <= k; x ++, y ++ ) q[y] = w[depth][x];
                for (x = l; x <= r; x ++, y ++ ) q[y] = w[depth][x];
                if (dfs(depth + 1, max_depth)) return true;   
                memcpy(q, w[depth], sizeof q);      //恢复现场
            }
    return false;
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T -- )
    {
        scanf("%d", &n);
        for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]);

        int depth = 0;
        //迭代加深
        while (depth < 5 && !dfs(0, depth)) depth ++ ;
        if (depth >= 5) puts("5 or more");
        else printf("%d\n", depth);
    }

    return 0;
}

 

181. 回转游戏

样例输入：

1 1 1 1 3 2 3 2 3 1 3 2 2 3 1 2 2 2 3 1 2 1 3 3
1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3
0

样例输出：

AC
2
DDHH
2

代码模板：

/*
      0     1
      2     3
4  5  6  7  8  9  10
      11    12
13 14 15 16 17 18 19
      20    21
      22    23
*/


#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 24;

int q[N];
int op[8][7] = {
    {0, 2, 6, 11, 15, 20, 22},
    {1, 3, 8, 12, 17, 21, 23},
    {10, 9, 8, 7, 6, 5, 4},
    {19, 18, 17, 16, 15, 14, 13},
    {23, 21, 17, 12, 8, 3, 1},
    {22, 20, 15, 11, 6, 2, 0},
    {13, 14, 15, 16, 17, 18, 19},
    {4, 5, 6, 7, 8, 9, 10}
};
int center[8] = {6, 7, 8, 11, 12, 15, 16, 17};
int opposite[8] = {5, 4, 7, 6, 1, 0, 3, 2};

int path[100];

int f()
{
    static int sum[4];
    memset(sum, 0, sizeof sum);
    for (int i = 0; i < 8; i ++ ) sum[q[center[i]]] ++ ;

    int s = 0;
    for (int i = 1; i <= 3; i ++ ) s = max(s, sum[i]);
    return 8 - s;
}

bool check()
{
    for (int i = 1; i < 8; i ++ )
        if (q[center[i]] != q[center[0]])
            return false;
    return true;
}

void operation(int x)
{
    int t = q[op[x][0]];
    for (int i = 0; i < 6; i ++ ) q[op[x][i]] = q[op[x][i + 1]];
    q[op[x][6]] = t;
}

bool dfs(int depth, int max_depth, int last)
{
    if (depth + f() > max_depth) return false;
    if (check()) return true;

    for (int i = 0; i < 8; i ++ )
    {
        if (opposite[i] == last) continue;
        operation(i);
        path[depth] = i;
        if (dfs(depth + 1, max_depth, i)) return true;
        operation(opposite[i]);
    }

    return false;
}

int main()
{
    while (scanf("%d", &q[0]), q[0])
    {
        for (int i = 1; i < N; i ++ ) scanf("%d", &q[i]);
        int depth = 0;
        
        while (!dfs(0, depth, -1)) depth ++ ;
        if (!depth) printf("No moves needed");
        for (int i = 0; i < depth; i ++ ) printf("%c", 'A' + path[i]);
        printf("\n%d\n", q[6]);
    }

    return 0;
}