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。

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X-Plosives

A secret service developed a new kind of explosive that attain its volatile property only when a specific association of products occurs. Each product is a mix of two different simple compounds, to which we call a binding pair. If N>2, then mixing N different binding pairs containing N simple compounds creates a powerful explosive. For example, the binding pairs A+B, B+C, A+C (three pairs, three compounds) result in an explosive, while A+B, B+C, A+D (three pairs, four compounds) does not.

You are not a secret agent but only a guy in a delivery agency with one dangerous problem: receive binding pairs in sequential order and place them in a cargo ship. However, you must avoid placing in the same room an explosive association. So, after placing a set of pairs, if you receive one pair that might produce an explosion with some of the pairs already in stock, you must refuse it, otherwise, you must accept it.

An example. Let’s assume you receive the following sequence: A+B, G+B, D+F, A+E, E+G, F+H. You would accept the first four pairs but then refuse E+G since it would be possible to make the following explosive with the previous pairs: A+B, G+B, A+E, E+G (4 pairs with 4 simple compounds). Finally, you would accept the last pair, F+H.

Compute the number of refusals given a sequence of binding pairs.

Input

Instead of letters we will use integers to represent compounds. The input contains several lines. Each line (except the last) consists of two integers (each integer lies between 0 and 10⁵) separated by a single space, representing a binding pair. The input ends in a line with the number –1. You may assume that no repeated binding pairs appears in the input.

Output

For each test case, a single line with the number of refusals.

Sample Input

1 2

3 4

3 5

3 1

2 3

4 1

2 6

6 5

-1

Sample Output

3

#include <stdio.h>
int p[100005], rank[100005];
int findp(int x) {
    return p[x] == x ? x : (p[x]=findp(p[x]));
}
int joint(int x, int y) {
    x = findp(x), y = findp(y);
    if(x == y)
        return 0;
    if(rank[x] > rank[y])
        rank[x] += rank[y], p[y] = x;
    else
        rank[y] += rank[x], p[x] = y;
    return 1;
}
int main() {
    int x, y, i;
    while(true) {
        if(scanf("%d %d", &x, &y) != 2)
            break;
        for(i = 0; i < 100005; i++)
            p[i] = i, rank[i] = 1;
        int ret = 0, flag;
        flag = joint(x, y);
        if(!flag)   ret++;
        while(scanf("%d", &x) == 1 && x >= 0) {
            scanf("%d", &y);
            flag = joint(x, y);
            if(!flag)   ret++;
        }
        printf("%d\n", ret);
    }
    return 0;
}