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 D: Forests

If a tree falls in the forest, and there's nobody there to hear, does it make a sound? This classic conundrum was coined by George Berkeley (1685-1753), the Bishop and influential Irish philosopher whose primary philosophical achievement is the advancement of what has come to be called subjective idealism. He wrote a number of works, of which the most widely-read are Treatise Concerning the Principles of Human Knowledge (1710) and Three Dialogues between Hylas and Philonous (1713) (Philonous, the "lover of the mind," representing Berkeley himself).

A forest contains T trees numbered from 1 to T and P people numbered from 1 to P.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

Standard input consists of a line containing P and T followed by several lines, containing a pair of integers i and j, indicating that person i has heard tree j fall. People may have different opinions as to which trees, according to Berkeley, have made a sound.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

How many different opinions are represented in the input? Two people hold the same opinion only if they hear exactly the same set of trees. You may assume that P < 100 and T < 100.

Sample Input

1

3 4
1 2
3 3
1 3
2 2
3 2
2 4

Output for Sample Input

2

---

暴力檢查是否有些人都看到了某些樹都倒了

#include <stdio.h>
#include <string.h>
int main() {
	int t, P, T;
	char blank[200];
	scanf("%d ", &t);
	while(t--) {
		scanf("%d %d ", &P, &T);
		int PT[101][101] = {};
		int i, j, k;
		while(gets(blank)) {
			if(strcmp(blank, "") == 0)
				break;
			sscanf(blank, "%d %d", &i, &j);
			PT[i][j] = 1;
		}
		int ans = 0, used[101] = {};
		for(i = 1; i <= P; i++) {
			if(used[i] == 0) {
				for(j = i+1; j <= P; j++) {
					if(used[j] == 0) {
						for(k = 1; k <= T; k++)
							if(PT[i][k] != PT[j][k])
								break;
						if(k == T+1)
							used[j] = 1;
					}
				}
				ans ++;
			}
		}
		printf("%d\n", ans);
		if(t)	puts("");
	}
    return 0;
}