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

Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others.

In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP).

However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

Input 

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n30000 and 0m500. Every student is numbered by a unique integer between 0 and n - 1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space.

A case wit n = 0 and m = 0 indicates the end of the input, and need not be processed.

Output 

For each case, output the number of suspects in one line.

Sample Input 

100 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0

Sample Output 

4
1
1

---

首先會是 0 當作感染的嫌疑人，接著與 0 相同社團的的人同樣也會當作嫌疑人，
而這些嫌疑人所在的社團的人同樣會是嫌疑人。
問嫌疑人共有多少？

很明顯地看出要找到一個連通元件，建圖的部分隨便找一個社團中的一人，將所有人連向她即可。

#include <stdio.h>
#include <vector>
using namespace std;
vector<int> g[30005];
int used[30005], ret;
void dfs(int x) {
    ret++, used[x] = 1;
    for(vector<int>::iterator it = g[x].begin();
        it != g[x].end(); it++) {
        if(used[*it] == 0) {
            dfs(*it);
        }
    }
}
int main() {
    int n, m, k;
    int i, j, x, y;
    while(scanf("%d %d", &n, &m) == 2 && n) {
        for(i = 0; i <= n; i++)
            g[i].clear(), used[i] = 0;
        while(m--) {
            scanf("%d", &k);
            scanf("%d", &x), k--;
            while(k--) {
                scanf("%d", &y);
                g[x].push_back(y);
                g[y].push_back(x);
            }
        }
        ret = 0;
        dfs(0);
        printf("%d\n", ret);
    }
    return 0;
}