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

A few kids are standing around an old tree playing a game. The tree is so huge that each kid can only see the kids close to him/her.

The game consists many �turns�. At the beginning of each turn of the game, a piece of paper is given to a randomly chosen kid. This kid writes the letter �B� if he is a boy or the letter �G� if a girl. Then he chooses a direction to pass the paper (clockwise or counter-clockwise), and gives the paper to his neighbor in that direction. The kid getting the paper writes down his sex too, and gives the paper to his neighbor in the same direction. In this way, the paper goes through the kids one by one, until one kid stops passing the paper and announces the end of this turn.

For example, there are five kids around the tree, and their genders are shown in Figure-1. The paper first goes to Kid1, after writing a �B� he passes it to Kid2, and Kid2 to Kid3. After Kid3 writes down a �G�, she ends up this turn, and we get the paper with a string �BBG�.

After N turns, we get N pieces of paper with strings of �B�s and/or �G�s. One of the kids will get all these papers, and has to figure out at least how many kids are around the tree playing the game. It's known that there are at least two kids. Please write a program to help him.

Input 

There are several test cases. Each case starts with a line containing an integer N, the number of papers (2 ≤ N ≤ 16). Each of the following N lines contains a string on a paper, which is a nonempty string of letter �B�s and/or �G�s. Each string has no more than 100 letters.

A test case of N = 0 indicates the end of input, and should not be processed.

Output 

For each test case, output the least possible number of kids in a line.

Sample Input 

3
BGGB
BGBGG
GGGBGB
2
BGGGBBBGG
GBBBG
0

Sample Output 

9
6

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[ICPC2004] 北京 Fun Game UVA1204, POJ2052, ZOJ2213
題目描述：

男女圍一桌，遊戲玩 n <= 16 輪。每輪隨機從一個人開始，指定順或逆時針傳，傳到的人寫下自己的性別，隨時都可能終止。給每輪得到的資訊，問最少可能有多少人。

題目討論：

討論相當少，POJ 上的討論也是跨年討論 orz。猜測相當於是 TSP 的狀態壓縮方式，O(2^16*16)。

只怕要處理單一組自繞資訊(縮繞幾圈的結果)，附帶一提這題 level 9，環狀處理相當重要，

處理的時候必然討論是一條鍊狀，而為了方便繞回來的檢查，
TSP 考慮時，從最長資訊開始討論，其後保證一定繞得回來。

如何保證每次連接最佳，最後一定會最佳？
考慮串接時，每組資訊的起頭位置必然呈現有序的，而且必然會在重疊後仍多出一小段，如果被包含前一段中，直接在預處理篩掉這種無須考慮的資訊。

由於每組資訊可能會存在順逆時針 O(2^16*16*2)。

先考慮單一資訊具有多圈的情況，使用 KMP 去進行檢測來加快速度。

狀態 O(2^16*16*2)，轉移次數共計 2^16*16*2*16 = 33554432。
使用 KMP 求 maximum suffix matching，否則建圖就需要 16*16*100*100 = 2560000。
使用 KMP 降至 16*16*100 = 25600。
建圖完之後，挑一個最長的出來當起點，使用 TSP 的 DP+bitmask。
如果答案等於最長的那個點，則開始進行自繞測試。
自繞測試最慘為 100*16*100*100 = 16000000。
用 KMP 降至 100*16*100 = 1600000。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
void kmpTable(char s[], int P[]) {
    int i, j;
    P[0] = -1, i = 1, j = -1;
    while(s[i]) {
        while(j >= 0 && s[j+1] != s[i])
            j = P[j];
        if(s[j+1] == s[i])
            j++;
        P[i++] = j;
    }
}
int suffixMatching(char A[], char B[], int la, int lb) { // B in A
	static int P[305];// modify KMP matching.
    int i, j, Alen, Blen;
    Alen = la, Blen = lb;
    kmpTable(B, P);
    for(i = 0, j = -1; i < Alen; i++) {
        while(j >= 0 && B[j+1] != A[i])
            j = P[j];
        if(B[j+1] == A[i])    j++;
        if(j == Blen-1)
            return j;
    }
    return j;
}
int cost[16][2][16][2];
int lapped[16][2][16][2];
int dp[1<<16][16][2], dp_st;
int dfs(int state, int last, int inv, int n) {
	int &ret = dp[state][last][inv];
	if(ret != -1)	return ret;
	int i;
	for(i = 0; i < n; i++) {
		if((state>>i)&1) {
			if(cost[last][inv][i][0] == 0 || cost[last][inv][i][1] == 0)
				state ^= (1<<i);
		}
	}	
	if(state == 0) {;
		if(dp_st != last)
			return -lapped[last][inv][dp_st][0];
		return 0;
	}
	ret = 0xfffffff;
	for(i = 0; i < n; i++) {
		if((state>>i)&1) {
			ret = min(ret, dfs(state^(1<<i), i, 0, n)+cost[last][inv][i][0]);
			ret = min(ret, dfs(state^(1<<i), i, 1, n)+cost[last][inv][i][1]);
		}
	}
	return ret;
}
int n;
int i, j, k;
int p, q, r;
char paper[16][2][255];
int paperlen[16];
int special_check() {
	char buf[305];
	int i, j;
	for(i = 1; i < paperlen[dp_st]; i++) {
		int mm = 2*paperlen[dp_st]+2;
		for(j = 0; j < mm; j++)
			buf[j] = paper[dp_st][0][j%(i+1)];
		buf[mm] = '\0';
		int ok = 1;
		for(j = 0; j < n && ok; j++) {
			int flag = 0, t;
			t = suffixMatching(buf, paper[j][0], mm, paperlen[j]);
			if(t == paperlen[j]-1)
				continue;
			t = suffixMatching(buf, paper[j][1], mm, paperlen[j]);
			if(t == paperlen[j]-1)
				continue;
			ok = 0;
		}
		if(ok)
			return i+1;
	}
	return 0xfffffff;
}
int main() {
	while(scanf("%d", &n) == 1 && n) {
		for(i = 0; i < n; i++) {
			scanf("%s", paper[i][0]);
			paperlen[i] = strlen(paper[i][0]);
			for(j = paperlen[i]-1, k = 0; j >= 0; j--, k++)
				paper[i][1][j] = paper[i][0][k];
			paper[i][1][paperlen[i]] = '\0';
		}
		for(i = 0; i < n; i++) {
			for(p = 0; p < 2; p++) {
				for(j = 0; j < n; j++) {
					for(q = 0; q < 2; q++) {
						int pos = suffixMatching(paper[i][p], paper[j][q], paperlen[i], paperlen[j]);
						cost[i][p][j][q] = paperlen[j]-pos-1;
						lapped[i][p][j][q] = pos+1;
					}
				}
			}
		}
		memset(dp, -1, sizeof(dp));
		dp_st = 0;
		for(i = 0; i < n; i++) {
			if(paperlen[i] > paperlen[dp_st])
				dp_st = i;
		}
		int ret = special_check();
		if(ret != 0xfffffff) {
			printf("%d\n", ret);
			continue;
		}
		ret = dfs((1<<n)-1-(1<<dp_st), dp_st, 0, n)+paperlen[dp_st];
		printf("%d\n", ret);
	}
	return 0;
}