Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

	Problem A	Arrange the Tiles	Time Limit : 4 seconds
	There is a board of dimension 4 x 3. Each cell of the board is a container that can hold a tile. The board is shown in the following diagram. As expected, you have got 12 tiles with you and you are required to place them in the containers so that every container has one tile in it. However, the tiles that you have are not ordinary tiles. Each tile is a square piece with all its 4 edges colored. A tile is described by its 4 edge colors starting from top and going clockwise. R G B Y R G R Y Y Y Y Y The image shows 3 tiles. The description of the tiles are given below the image. The colors that will be used to denote the tiles will be from the set {yellow, green, blue and red} and will be represented by {Y, G, B and R} for the sake of brevity. There are 12 factorials (12!) ways of placing the tiles on the board. A placement is considered fragile if the touching sides of any two adjacent tiles is made up of different colors. You are required to find out the total number of placements which are not fragile. Note: You can not rotate the tiles. That is, the initial orientation must be preserved.
	Input
	The first line of input is an integer T( T < 20 ) that denotes the number of test cases. Each case starts on a new line and consists of one or more lines containing 12 strings. Each string represents a tile and is made up of 4 characters.
	Output
	For each case, output the case number followed by the total number of non-fragile placements.
	Sample Input	Sample Output
	2 BBBB BBBB BBBB BBBB BBBB BBBB BBBB BBBB BBBB BBBB BBBB YYYY GGGG GGGG GGGG GGGG GGGG GGGG GGGG GGGG GGGG GGGG GGGG GGGG	Case 1: 0 Case 2: 479001600
	Problem Setter: Sohel Hafiz Special Thanks: Shamim Hafiz, Md. Arifuzzaman Arif Next Generation Contest 5

題目描述：

排列 12 塊的方塊成 4x3 如圖那樣，每塊最多使用一次，因此是 12! 種，而每塊四邊有不同的顏色，

放上去時，相鄰的邊要共同色，問總共有多少種排法。

題目解法：

拆成一半 2x3，考慮將這一半的所有可能找出來，並且記錄這個 2x3 最上邊的 3 個顏色和下邊的 3 個顏色，因此會有 C(12, 6) = 924 種，顏色則會有 4^3 = 64。

考慮將使用過的排列，以 bitmask 記錄，以方便我們進行兩個 2x3 合併。

合併的時候很簡單，下邊顏色要等於上邊顏色，而且使用的塊會互補，方法就是相乘起來。


#include <stdio.h>
#include <string.h>
char tile[12][10];
int dpup[64][4096], dpdown[64][4096];
// up side&down side, 4^3 = 64, each state 2^12 = 4096
int chways[6], used[12] = {};//chose 6 item
int panel[2][3];
int trans(char c) {
    switch(c) {
        case 'R':return 0;
        case 'G':return 1;
        case 'B':return 2;
        case 'Y':return 3;
    }
    puts("err");
    return -1;
}
void dfs(int idx) {
    int i;
    int x, y;
    if(idx == 6) {
        int up = 0, down = 0, state = 0;
        int d;
        for(i = 0; i < 6; i++)
            state |= 1<<chways[i];
        for(i = 0; i < 3; i++)
            up = up*4 + trans(tile[panel[0][i]][0]);
        for(i = 0; i < 3; i++) {
            down = down*4 + trans(tile[panel[1][i]][2]);
        }
        dpup[up][state]++;
        dpdown[down][state]++;
        return;
    }
    for(i = 0; i < 12; i++) {
        if(used[i] == 0) {
            x = idx/3, y = idx%3;
            if(x-1 >= 0) {// up
                if(tile[panel[x-1][y]][2] != tile[i][0]) {
                    continue;
                }
            }
            if(y-1 >= 0) {
                if(tile[panel[x][y-1]][1] != tile[i][3]) {
                    continue;
                }
            }
            chways[idx] = i;
            panel[x][y] = i;
            used[i] = 1;
            dfs(idx+1);
            used[i] = 0;
        }
    }
}
int main() {
    int testcase, cases = 0;
    int i, j, k;
    scanf("%d", &testcase);
    while(testcase--) {
        for(i = 0; i < 12; i++)
            scanf("%s", tile[i]);
        memset(dpup, 0, sizeof(dpup));
        memset(dpdown, 0, sizeof(dpdown));
        dfs(0);
        long long ret = 0;
        for(i = 0; i < 64; i++) {
            for(j = 0; j < 4096; j++) {
                ret += dpdown[i][j]*dpup[i][4095-j];
            }
        }
        printf("Case %d: %lld\n", ++cases, ret);
    }
    return 0;
}