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

 Word Transformation 

A common word puzzle found in many newspapers and magazines is the word transformation. By taking a starting word and successively altering a single letter to make a new word, one can build a sequence of words which changes the original word to a given end word. For instance, the word ``spice'' can be transformed in four steps to the word ``stock'' according to the following sequence: spice, slice, slick, stick, stock. Each successive word differs from the previous word in only a single character position while the word length remains the same.

Given a dictionary of words from which to make transformations, plus a list of starting and ending words, your team is to write a program to determine the number of steps in the shortest possible transformation.

Input and Output

The first line of the input is an integer N, indicating the number of test set that your correct program should test followed by a blank line. Each test set will have two sections. The first section will be the dictionary of available words with one word per line, terminated by a line containing an asterisk (*) rather than a word. There can be up to 200 words in the dictionary; all words will be alphabetic and in lower case, and no word will be longer than ten characters. Words can appear in the dictionary in any order.

Following the dictionary are pairs of words, one pair per line, with the words in the pair separated by a single space. These pairs represent the starting and ending words in a transformation. All pairs are guaranteed to have a transformation using the dictionary given. The starting and ending words will appear in the dictionary.

Two consecutive input set will separated by a blank line.

The output should contain one line per word pair for each test set, and must include the starting word, the ending word, and the number of steps in the shortest possible transformation, separated by single spaces. Two consecutive output set will be separated by a blank line.

Sample Input

1

dip
lip
mad
map
maple
may
pad
pip
pod
pop
sap
sip
slice
slick
spice
stick
stock
*
spice stock
may pod

Sample Output

spice stock 4
may pod 3

---

作法 : 判斷任兩字串的轉換最短路徑長, 以下用 BFS

#include<stdio.h>
#include<string.h>
int main() {
	char s[101], sx[101], sy[101], D[201][20];
	int map[201][201], mt[201], size;
	int n, i, j, k, fx, fy;
	scanf("%d", &n);
	getchar();
	gets(s);
	while(n--) {
		size = 0;
		while(gets(s)) {
			if(s[0] == '*')	break;
			strcpy(D[size], s);
			size++;
		} 
		memset(mt, 0, sizeof(mt));
		for(i = 0; i < size; i++) {
			for(j = 0; j < size; j++) {
				int dif = 0;
				for(k = 0; D[i][k] && D[j][k]; k++)
					if(D[i][k] != D[j][k])
						dif++;
				if(D[i][k] == 0 && D[j][k] == 0 && dif == 1)
					map[i][mt[i]++] = j, map[j][mt[j]++] = i;
			}
		}
		while(gets(s)) {
			if(s[0] == '\0')	break;
			sscanf(s, "%s %s", sx, sy);
			for(i = 0; i < size; i++)
				if(!strcmp(sx, D[i])) {
					fx = i;break;
				}
			for(i = 0; i < size; i++)
				if(!strcmp(sy, D[i])) {
					fy = i;break;
				}
			printf("%s %s ", sx, sy);
			if(fx == fy)	puts("0");
			else {
				int Used[size], Q[size*size], Qt = -1, V[size], tn;
				memset(Used, 0, sizeof(Used));
				Q[++Qt] = fx, V[fx] = 0, Used[fx] = 1;
				for(i = 0; i <= Qt; i++) {
					tn = Q[i];
					if(tn == fy)	break;
					for(j = 0; j < mt[tn]; j++)
						if(Used[map[tn][j]] == 0) {
							Used[map[tn][j]] = 1, V[map[tn][j]] = V[tn]+1;
							Q[++Qt] = map[tn][j];
						}
				}
				printf("%d\n", V[fy]);
			}
		}
		if(n)	puts("");
	}
    return 0;
}