Morris' Blog 我在摸索我存在的意、生命的意、涵及值。 有真正神手般的害，套用模式是大部分的情。 那害的我存在的意何，托出神的存在？ 有候不能怪人不我，我已始省思， 什我有人的原因，而是全要靠人。 有候我我不同，那是因我做不到很多人能到的事情。 我不到、玩不起、得不明、理解得不多， 我缺少的文能力，就是一切一切能的判定。 此，我不得能做些人有助的事情。 我知道我自己是不服的性格，但在我不得不。 我有能力，因此我必。 我到底世界作什？充不定性， 而我是法出拔萃，想一般人似乎也回不了了。 越大的挑需要越力的支持，而我的支持是什？ 最近人，但都是失的局，人能接受， 不需要言，不需要通，一切所想都表不出。

49的鼓 6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

目描述：

排列 12 的方成 4x3 如那，每最多使用一次，因此是 12! ，而每四有不同的色，

放上去，相的要共同色，共有多少排法。

目解法：

拆成一半 2x3，考一半的所有可能找出，且 2x3 最上的 3 色和下的 3 色，因此有 C(12, 6) = 924 ，色有 4^3 = 64。

考使用的排列，以 bitmask ，以方便我行 2x3 合。

合的候很，下色要等於上色，而且使用的互，方法就是相乘起。


#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;
}

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