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

One of the reasons for which École polytechnique (nicknamed ``X" for reasons to be explained during the debriefing talk) is so deeply rooted in French society is its famous network of camarades - former students of the same school. When one camarade wants something (money, job, etc.), he can ask this network for help and support. In practice, this means that when he/she wants to reach some other camarade, not always of the same year, then surely he can find intermediate camarades to get to her/him. Note that the ``camarade" relationship is symmetric. Due to the magic of the X network, there is always a means to reach anybody.

The program you have to write is supposed to help to minimize the number of these intermediate camarades.

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.

The huge file of all living camarades is simplified so as to obey the following format. The first line in the file is the number of camarades, say N , an integer 1N10⁵ . Camarades are labeled from 0 to N - 1 . Follow N lines. Each line starts with the camarade label c , followed by the number of other camarades he/she knows, say n_c , followed by the labels of those n_c camarades. All these integers are separated by a single blank. It is assumed that n_c is always less than 100. The last line in the file is the label of the camarade seeking help (say c₁ ) followed by the label of the camarade he wants help from, say c₂ ( c₂ c₁ ).

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.

Your program should output three integers separated by a blank: c₁ , c₂ and the minimal number of intermediate camarades to reach c₂ .

Sample Input 

1

4
0 3 1 2 3
1 1 0
2 2 0 3
3 2 0 2
1 2

Sample Output 

1 2 1

---

題目描述：

想要尋求某人的幫助，最少要轉接幾個人的手。

題目解法：

事實上只是單純地找到最短路。權重都是 1，直接 BFS。

#include <stdio.h>
#include <vector>
#include <queue>
using namespace std;
vector<int> g[100005];
inline int ReadInt(int *x) {
    static char c, neg;
    while((c = getchar()) < '-')    {if(c == EOF) return EOF;}
    neg = (c == '-') ? -1 : 1;
    *x = (neg == 1) ? c-'0' : 0;
    while((c = getchar()) >= '0')
        *x = (*x << 3) + (*x << 1) + c-'0';
    *x *= neg;
    return 1;
}
int main() {
    int testcase;
    int n, m, x, y;
    int i, j, k, st, ed;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d", &n);
        for(i = 0; i < n; i++)
            g[i].clear();
        for(i = 0; i < n; i++) {
            //scanf("%d %d", &x, &m);
            ReadInt(&x);
            ReadInt(&m);
            while(m--) {
                //scanf("%d", &y);
                ReadInt(&y);
                g[x].push_back(y);
            }
        }
        scanf("%d %d", &st, &ed);
        queue<int> Q;
        int dist[100005] = {};
        Q.push(st);
        dist[st] = 1;
        while(!Q.empty()) {
            x = Q.front(), Q.pop();
            for(i = 0; i < g[x].size(); i++) {
                if(dist[g[x][i]] == 0) {
                    dist[g[x][i]] = dist[x]+1;
                    Q.push(g[x][i]);
                }
            }
            if(dist[ed])    break;
        }
        printf("%d %d %d\n", st, ed, dist[ed]-2);
        if(testcase)
            puts("");
    }
    return 0;
}