剪枝 ... ++ 全局 %d figure sym ati 情況

Description

The rotation game uses a # shaped board, which can hold 24 pieces of square blocks (see Fig.1). The blocks are marked with symbols 1, 2 and 3, with exactly 8 pieces of each kind.

Initially, the blocks are placed on the board randomly. Your task is to move the blocks so that the eight blocks placed in the center square have the same symbol marked. There is only one type of valid move, which is to rotate one of the four lines, each consisting of seven blocks. That is, six blocks in the line are moved towards the head by one block and the head block is moved to the end of the line. The eight possible moves are marked with capital letters A to H. Figure 1 illustrates two consecutive moves, move A and move C from some initial configuration.

Input

The input consists of no more than 30 test cases. Each test case has only one line that contains 24 numbers, which are the symbols of the blocks in the initial configuration. The rows of blocks are listed from top to bottom. For each row the blocks are listed from left to right. The numbers are separated by spaces. For example, the first test case in the sample input corresponds to the initial configuration in Fig.1. There are no blank lines between cases. There is a line containing a single `0‘ after the last test case that ends the input.

Output

For each test case, you must output two lines. The first line contains all the moves needed to reach the final configuration. Each move is a letter, ranging from `A‘ to `H‘, and there should not be any spaces between the letters in the line. If no moves are needed, output `No moves needed‘ instead. In the second line, you must output the symbol of the blocks in the center square after these moves. If there are several possible solutions, you must output the one that uses the least number of moves. If there is still more than one possible solution, you must output the solution that is smallest in dictionary order for the letters of the moves. There is no need to output blank lines between cases.

Sample Input

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

Sample Output

AC
2
DDHH
2

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日常吐槽：調了半天玄學錯誤，最後發現我遞歸的時候修改了全局變量的值...這個錯誤調了我大概一個小時...
哎還是太菜了，要加油啊。
然後過了樣例提交，WA，改了改某些自己認為不太對的地方，WA，看了討論，發現沒有在不用移動的情況下輸出第二行，改了，WA，重看，發現沒刪調試語句。
AC！！！！！547Ms感覺不錯。
這題用了我大概兩個小時。
算法：搜索。
搜索框架：dfs(顯然)。
但是這樣是顯然不能過的。
剪枝思路：因為可能的步數不是很多，所以可以考慮叠代加深A*，下面的問題是我們如何確定估價函數。
觀察到題目特點，每次操作最多只能增加中間的一個數字，如果中間的格子數字最多出現次數是cnt，那麽至少移動8-cnt步才能到達，可以想象這個估價函數非常強(霧)。
然後獻上自己醜陋的代碼。
也紀念一下我的第一道IDA*的題目。

---

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define reg register
inline int read() {
    int res=0;char ch=getchar();bool fu=0;
    while(!isdigit(ch))fu|=(ch==‘-‘), ch=getchar();
    while(isdigit(ch))res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
    return fu?-res:res;
}
int a[8][8];
int ans, road[105];
bool Flag;
int mark;

void IDAstar(int x[8][8], int dep) {
//    printf("%d::\n", dep);
//    for (int i=1;i<=7;i++,puts("")) for(int j=1;j<=7;j++) if(!a[i][j]) printf("  ");else printf("%d ",x[i][j]);puts("");
    int b[8][8];
    if (Flag) return;
    int mx = 1;
    int cnt[4] = {0};
    for (reg int i = 3 ; i <= 5 ; i ++) for (reg int j = 3 ; j <= 5 ; j ++) cnt[x[i][j]]++;
    for (reg int i = 2 ; i <= 3 ; i ++) if (cnt[i] > cnt[mx]) mx = i;
    if (dep == ans) {
        if (cnt[mx] == 8) {
            Flag = 1, mark = mx;
            for (reg int i = 0 ; i < dep ; i ++) printf("%c", road[i] + ‘A‘ - 1);
            printf("\n%d\n", mx);
        }
//        for (int i=1;i<=3;i++) printf("%d ", cnt[i]);puts("");
        return ;
    }
    if (dep + 8 - cnt[mx] > ans) return;
    
    // A:
    memcpy(b, x, sizeof b);
    for (reg int i = 1 ; i < 7 ; i ++) b[i][3] = x[i + 1][3];
    b[7][3] = x[1][3];
    road[dep] = 1;
    IDAstar(b, dep + 1);
    // B:
    memcpy(b, x, sizeof b);
    for (reg int i = 1 ; i < 7 ; i ++) b[i][5] = x[i + 1][5];
    b[7][5] = x[1][5];
    road[dep] = 2;
    IDAstar(b, dep + 1);
    // C:
    memcpy(b, x, sizeof b);
    for (reg int i = 7 ; i > 1 ; i --) b[3][i] = x[3][i - 1];
    b[3][1] = x[3][7];
    road[dep] = 3;
    IDAstar(b, dep + 1);
    
    // D:
    memcpy(b, x, sizeof b);
    for (reg int i = 7 ; i > 1 ; i --) b[5][i] = x[5][i - 1];
    b[5][1] = x[5][7];
    road[dep] = 4;
    IDAstar(b, dep + 1);
    // E:
    memcpy(b, x, sizeof b);
    for (reg int i = 7 ; i > 1 ; i --) b[i][5] = x[i - 1][5];
    b[1][5] = x[7][5];
    road[dep] = 5;
    IDAstar(b, dep + 1);
    // F:
    memcpy(b, x, sizeof b);
    for (reg int i = 7 ; i > 1 ; i --) b[i][3] = x[i - 1][3];
    b[1][3] = x[7][3];
    road[dep] = 6;
    IDAstar(b, dep + 1);    
    // G:
    memcpy(b, x, sizeof b);
    for (reg int i = 1 ; i < 7 ; i ++) b[5][i] = x[5][i + 1];
    b[5][7] = x[5][1];
    road[dep] = 7;
    IDAstar(b, dep + 1);
    // H:
    memcpy(b, x, sizeof b);
    for (reg int i = 1 ; i < 7 ; i ++) b[3][i] = x[3][i + 1];
    b[3][7] = x[3][1];
    road[dep] = 8;
    IDAstar(b, dep + 1);
}

int main()
{
    while(1) 
    {
        a[1][3] = read();if (!a[1][3]) return 0;
        a[1][5] = read();
        a[2][3] = read(), a[2][5] = read();
        for (reg int i = 1 ; i <= 7 ; i ++) a[3][i] = read();
        a[4][3] = read(), a[4][5] = read();
        for (reg int i = 1 ; i <= 7 ; i ++) a[5][i] = read();
        a[6][3] = read(), a[6][5] = read();
        a[7][3] = read(), a[7][5] = read();
//        for (int i=1;i<=7;i++,puts("")) for(int j=1;j<=7;j++) if(!a[i][j]) printf("  ");else printf("%d ",a[i][j]);
        int mx = 1;
        int cnt[4] = {0};
        for (reg int i = 3 ; i <= 5 ; i ++) for (reg int j = 3 ; j <= 5 ; j ++) cnt[a[i][j]]++;
        for (reg int i = 2 ; i <= 3 ; i ++) if (cnt[i] > cnt[mx]) mx = i;
        if (cnt[mx] == 8) {printf("No moves needed\n");printf("%d\n", mx);continue;}
        
        ans = 1;
        Flag = 0;
        for ( ; !Flag ; IDAstar(a, 0), ans ++);
        
//        for (reg int i = 1 ; i <= ans ; i ++) printf("%d", road[i]);
//        puts("");printf("%d\n", mark);
    }
    return 0;
}

[POJ2286]The Rotation Game